This problem (with slightly different wording) had appeared here in this forum, as being given at some TST. I commented it in my blog without knowing it's from ISL 2019.

Wow, I apparently have solved an A3. Hope it isn't a fakesolve! Firstly, observe that the condition of $ \left|1-\sum_{i \in X} a_{i}\right| = \delta$ being minimal is equivalent to $$ \frac{\left|\sum_{i \in X^C} a_i -\sum_{i \in X} a_{i}\right|}{2} $$be minimal. Then, assume WLOG that $\sum_{i \in X^C} a_i > \sum_{i \in X} a_{i}$ (if the equality ocurred, we would be done). Also, for any strictly increasing sequence of $n$ positive real numbers $(c_1,c_2,...,c_n)$, let $$\Delta_{(c_1,c_2,...,c_n)}= \left|\sum_{i \in X^C} c_i -\sum_{i \in X} c_{i}\right|$$We want to prove that there exists a strictly increasing sequence of $n$ positive real numbers $(b_1,b_2,...,b_n)$ such that $$\Delta_{(b_1,b_2,...,b_n)}=0$$ Now, observe that since $\frac{\Delta_{(a_1,a_2,...,a_n)}}{2}$ is minimal, for all $a_y \in X^C$ and $a_x \in X$ (they exist, since $\Delta_{(a_1,a_2,...,a_n)}>0$, we have that $$\frac{\Delta_{(a_1,a_2,...,a_n)}}{2} < a_y-a_x \qquad(\star)$$otherwise we could place $a_y$ in $X$ and $a_x$ in $X^C$ and $\frac{\Delta_{(a_1,a_2,...,a_n)}}{2}$ would decrease and would be greater than $0$, by assumption, and we would have a better choice for $X$, a contradiction! Then, take $x_0 \in X, y_0 \in X^C$ with $a_{x_0} < a_{y_0}$ such that $a_{y_0}-a_{x_0}=M$ is minimal. From $(\star)$, $ \delta= \frac{\Delta_{(a_1,a_2,...,a_n)}}{2} < M$. Thus, if we choose $b_i=a_i+\frac{\delta}{|X|}$ for all $i \in X$ and $b_i=a_i-\frac{\delta}{|X^C|}$, the order of the terms in $b_i \in X$ is the same of $a_i \in X$, as well as the order of the terms in $b_i \in X^C$ is the same of $a_i \in X^C$. Furthermore, $b_y-b_x < M$, from $(\star)$, so we don't have issues with an index $i \in X$ sweeping with and index $j \in X^C$ in the new choice of sequence (the sequence keeps strictly increasing). Hence, this new sequence satisfies $\Delta_{(b_1,b_2,...,b_n)}=0$, $\sum_{i=1}^n b_i = \sum_{i=1}^n a_i = 2$ and $\sum_{i \in X} b_{i}=1$ by construction. Therefore, we are done. $\blacksquare$

We solve the problem by characterizing all sets $X$ which can occur. To this end, write $Y=X^c$ for the complement and $X>Y$ ("$X$ majorizes $Y$") if for every $k$, the set $X$ contains at least as many indices larger than $k$ as does $Y$. Similarly, define $Y>X$. (Note that we do not at all need to have either $X>Y$ or $Y>X$.) Now if $X>Y$ it is clear that $\sum_{i \in X} b_i>1$. Similarly, if $Y>X$ it is clear that $\sum_{i \in X} b_i<1$. So we better prove that such sets $X$ cannot occur. Indeed, let us prove the following two facts: 1) If $X>Y$, then $X$ can not be a minimizing set as in the problem statement. 2) If we do not have $X>Y$, then we can find $b_i$ such that $\sum_{i \in X} b_i<1$. Before we prove 1) and 2), let us see how this finishes the problem. From 1) we get that the set $X$ from the problem must have neither $X>Y$ nor $Y>X$. Then from 2) we get a sequence $b_i$ such that $\sum_{i \in X} b_i<1$ and a sequence $c_i$ such that $\sum_{i \in X} c_i>1$. Taking a convex linear combination (which will still be strictly increasing and have total sum equal to $2$) we find a sequence with sum over $X$ exactly $1$. Finally, let us see how to prove 1) and 2): Proof of 1): Indeed, write $X=\{x_1,\dots,x_k\}$ and $Y=\{y_1,\dots,y_m\}$ with the elements in decreasing order. So $x_1 >y_1, x_2 > y_2$ etc. So $\sum_{i \in Y} a_i<1<\sum_{i \in Y} x_i$. But if $k \ge 2$ and $m \ge 1$ the set $M=\{y_1,x_2,\dots,x_k\}$ will have a sum strictly between the two sums, hence $X$ can not be a minimizing set. So we must have $X=\{n\}$ or $X=\{1,2,\dots,n\}$ both of which are easily seen to be only possibly minimizing if $n \le 2$ (so this is where our proof uses $n \ge 3$). Done. Proof of 2): So for some $k$ the set $X$ contains less than half of the indices from $\{k+1,\dots,n\}$. Choose $c_i=0$ for $i \le k$ and $c_i=\frac{2}{n-k}$ for $i>k$. This is an increasing sequence with $\sum_{i \in X} c_i<1$. Now just perturb $c_i$ to a sequence $b_i$ by a sufficiently small $\varepsilon$ to make it strictly increasing and still have $\sum_{i \in X} b_i<1$. Done.

Ok. I realize that this problem isn't particularly hard; it's just really annoying. Split $a_1, \ldots , a_n$ into $X = \{x_1, \ldots , x_k\}$ and $Y = \{y_1, \ldots , y_{\ell}\}$ where $k + \ell = n$. Denote the sum of all $x_i$ as $1 + \epsilon$ and the sum of all $y_i$ as $1 - \epsilon$. If $\epsilon$ is ever $0$, then we are done. So WLOG $\epsilon > 0$. Note that $\epsilon \leq x_i - y_j$ for all $i, j$ (where $x_i > y_j$ at least). If otherwise, then we may send $y_j$ to $X$ and $x_i$ to $Y$ and obtain a smaller $\epsilon '$ since\[1 - \epsilon < \sum Y' = (1 - \epsilon) + (x_i - y_j) \leq 1.\]In fact, we may further assume $\epsilon < x_i - y_j$ since if equality occurs, then after the swap, $\epsilon ' = 0$. Similarly, with this swapping argument, we obtain that all $x_i$ are greater than $\epsilon$. Now we construct a sequence $b_1, \ldots , b_n$ that satisfies the requirements. Let $S$ be the set of indices $i$ for which $a_i$ is a $x$ term and $T$ be the set of indices $j$ for which $a_j$ is a $y$ term. Clearly $|S| = k$ and $|T| = \ell$. I claim that the following sequence, defined by\begin{align*}b_i = a_i - \tfrac{\epsilon}{k} \text{ }\forall i \in S\\b_i = a_i + \tfrac{\epsilon}{\ell} \text{ }\forall i \in T\end{align*}works. Indeed, check that $\sum b_i = \sum a_i + {\epsilon} - {\epsilon} = 2$. and $\sum_{i \in S} b_i = (1 + \epsilon) - \epsilon = 1$. It remains to check that this sequence is strictly increasing. If $b_m, b_{m+1}$ are either both in $S$ or both in $T$, then it clearly follows that $b_m < b_{m+1}$ since $a_m < a_{m+1}$. If $b_m \in S$ and $b_{m+1} \in T$, then also $b_{m+1} - b_m = (a_{m+1} - a_m) + \tfrac{\epsilon}{k} + \tfrac{\epsilon}{\ell} > 0$. Now, if $b_m \in T$ and $b_{m+1} \in S$, then\[b_{m+1} - b_m = (x_i - y_j) - (\tfrac{\epsilon}{k} + \tfrac{\epsilon}{\ell}) \geq (x_i - y_j) - \epsilon > 0\]by our first note, along with using the fact that $\tfrac{1}{k} + \tfrac{1}{\ell} \leq \tfrac{1}{2} + \tfrac{1}{2n - 4} \leq 1$ for $n \geq 3$ unless $k$ or $\ell$ is $1$. If $k = 1$, then the only $x$ term has value $1 + \epsilon > 1$ which is bad since then the next largest term is a $y$-term and has value $< 1$, contradicting minimality of $\epsilon$. If $\ell = 1$, then the only $y$ term has value $1 - \epsilon$. The sum of the $n - 1$ remaning terms is $1 + \epsilon$. The sum of the smallest $n - 2$ terms is $> (n-2)\epsilon$, so the largest $x$ term is $< 1 + (3 - n)\epsilon = 1$. So if $x_{max} >$ the only $y$ term, we get a contradiction since their difference is $< \epsilon$. Thus, the $y$ term is the largest $a_i$, meaning that $b_m \in T$ and $b_{m+1} \in S$ is impossible. So indeed, this sequence $b_i$ works. $\blacksquare$

"No way an A3 could be THAT hard ..." $\Rightarrow$ me during the test, when I finally did it for 2.5x the average time other TST takers needed. Set-theory oriented solution in a completely different flavor: (first idea akin to the third solution from the official Shortlist). For notational purposes let $[n]$ be the set $\{ 1,2,\ldots,n\}$. Denote $a_{i+1} - a_i = d_i$, and $a_1$ be $d_0$. Note that the objective is equivalent to finding a sequence $\{b_i, 1 \leq i \leq n\}$ so that \[ \sum_{i\in X} b_i = \sum_{j \in [n]-X} b_j \]and also that the problem statement is equivalent to $X$ satisfying \[ \left| \sum_{i\in X} a_i - \sum_{j \in [n]-X} a_j \right| \]being the lowest possible among all set $X \subseteq [n]$, with the real numbrers $a_i$ fixed. $\textcolor{green}{\textbf{\text{Main idea.}}}$ Bash for the win! We directly compute the terms \[ S_d = \sum_{j\in Y} a_j - \sum_{i \in X} a_i = \sum_{i = 0}^{n-1} c_id_i \]in terms of $d_i$ (here, we let $Y = [n] - X$ for simplicity). Without Loss of Generality we can assume $X$ to be the set with the smaller sum. \[ \begin{tabular}{cccccccccccc} $a_i|\,\, i\in X$=&0&&$a_1$&&&&$a_3$\\ $a_j|\,\, j\in Y$=&&&&&$a_2$&&&&$a_4$&&$a_5$ \\ &&\kern-2\tabcolsep\upbracefill\kern-2\tabcolsep&&\kern-2\tabcolsep\upbracefill\kern-2\tabcolsep&&\kern-2\tabcolsep\upbracefill\kern-2\tabcolsep&&\kern-2\tabcolsep\upbracefill\kern-2\tabcolsep&&\kern-2\tabcolsep\upbracefill\kern-2\tabcolsep \\ &&$d_0$&&$d_1$&&$d_2$&&$d_3$&&$d_4$ \\[3pt] &&&&&&&&&&\text{Y-block} \end{tabular} \]For example, when $X = \{1,3\}$ and $Y = \{2,4,5\}$, $S_d$ (the letters $S_d$ represents the sum of "differences" $d_i$) = $1 \cdot d_0+ 2 \cdot d_1 + 1 \cdot d_2 + 2 \cdot d_3 + 1 \cdot d_4$. Note that if not all coefficients $c_i$ is of "unified" sign, e.g. there exists a positive and negative coefficient, we can $\textbf{easily}$ set the $d_i$ so that \[ \sum_{i; c_i > 0} |c_i|d_i = \sum_{j; c_j<0} |c_j|d_j \](one way to do it is to set all $d_i$ in the $LHS$ to be $1$, all $d_j$ except one on the $RHS$ to be $\frac{0.001}{n^2}$ and adjust the one not yet filled accordingly.) In these cases, setting $b_i$ equal to the new $d_0+\cdots + d_{i-1}$ fulfills the objective. So let all $c_i$ be positive or $0$, as these are the exceptional cases where we would like to investigate further (and ultimately, prove that these cases do not exist by the statement given in the problem.) Before we continue, we present a simple, but central observation. $\textcolor{blue}{\textbf{\text{Observation.}}}$ Let $i \in X$ and $i+1 \in Y$. If $X' = X \cup \{i+1\} - \{i\}$, then $S_d' = S_d - 2d_i$. $\textcolor{blue}{\textbf{\text{Proof.}}}$ Obvious [Just note that $S_d' = S_d - (a_{i+1} - a_{i}) + (a_{i} - a_{i+1})].$ $\rule{25 cm}{3pt}$ We have two different finishes from here. $\textcolor{red}{\textbf{\text{Finish 1: Polished, after redoing the problem.}}}$ Divide the indices into $\textit{blocks}$ of $X$ and $Y$. For example, in the set $X = \{1,4,6\}$ and $Y = \{2,3,5,7\}$, set the $Y-$blocks to be $\{2,3\}, \{5\}$ and $\{7\}$. Define an $X-Y$ transitional index to be the last index in any $X$- block, and say the $X-Y$ transitional index of an $X-$block to be its final index. (this is another way of categorizing the indices in which we can make a switch in the above $\textcolor{blue}{\textbf{\text{Observation.}}}$). We claim the following: $\textbf{Claim 1.}$ Each $X-Y$ transitional index $i$ has coefficient $c_i$ at least $1$. $\textbf{Proof 1.}$ Consider the coefficient $c_{i+1}$ of $d_{i+1}$. Then, we can be sure that $c_{i} = c_{i+1} +1 \geq 0+1 = 1$. Here we use the observation that $c_i$ is the difference between "the amount of indices in $Y$" and "the amount of indices in $X$" greater than $i$. $\textbf{Claim 2.}$ There are at most one $X-Y$ transitional index. $\textbf{Proof 2.}$ If $i$ and $j$ are $X-Y$ transitional indices, then \[ S_d(X) \geq c_id_i + c_jd_j \geq 1 \cdot d_i + 1 \cdot d_j \]Consider $X'(i)$ and $X'(j)$ after switching $i$ with $i+1$, or $j$ with $j+1$ respectively. Then, one of $S_d(X'(i)) \geq d_i - d_j$ and $S_d(X'(j)) \geq d_j-d_i$ is at least $0$, but less than $S_d$. $\rule{25cm}{0.5pt}$ These two claims greatly delimit the possible configurations of $X$, as we can just let $X$ to be an interval $[i,j] \in [n]$ or a collection of two intervals $[a,b], [c,d]$ where $0<a<b<c<d = n$. However, the second case can also be easily ruled out as $X$ does not contain the largest number $a_n$. On the first case, if $|X| > n-j$; i.e. there are more terms in $X$ rather than the ending $Y-$block, then the coefficient of $d_{i-1}$ is negative. Otherwise we can be sure that the $X-$ block is kind of small. First subcase: if the ending $Y-$ block's size is $2$ or greater: then we know that $d_j$ is at least 2. Then consider $X'$ obtaining from switching $j$ with $j+1$. Second subcase: ending $Y-$ block's size is $1$. Then, the size of the $X-$ block is also 1, and a little experimentation yields that choosing $X = \{n\}$ contradicts the $\textbf{original}$ problem statement. (While the value of $a_n$ alone can certainly be BIG, it cannot be BIG enough to overrun the initial difference) (Note: configurations with no $X-Y$ transitional indices may be dealt with in the same way as post $\#2$ does it) $\rule{25 cm}{3pt}$ $\textcolor{red}{\textbf{\text{Finish 2: Original 3 hour solution "Y-block maniac".}}}$ We refer to $Y-X$ transitional indices to be the last index in a (particular) $Y-$ block. We claim the following: $\textbf{Claim 3.}$ The length of any $Y-$ block aside from the first one is exactly $1$. $\textbf{Proof 3.}$ Let a $Y-$ block has length $l \geq 2$. Consider the transitional index of that block $t$. Now, using a similar principle to $\textbf{Claim 1}$ above we get $c_{t-l} = c_t+l \geq 0+l = l \geq 2$. Letting $X'$ with switching indices $t-l$ and $t-l+1$ and using the initial $\textcolor{blue}{\textbf{\text{Observation}}}$, a contradiction takes place. $\rule{25cm}{0.5pt}$ This claim also delimits the possibilities of $X-$ blocks preceding each $Y-$ block, as each block must have length at least $1$. It is illogical for any $X-$ block to have length $2$ or more: as the leftmost coefficient of said block will inevitably be negative. If we have $2$ or more $X-$ blocks, we now pick two which are $\{i\}$ and $\{j\}$. Let the index with the smaller "difference" be $i$ (i.e. $d_i \leq d_j$). Then there is an easy switch of index $i$ and $i+1$ which yields a contradiction. The case where the only $X-$ block available is of length $1$ is handled similarly to $\textcolor{red}{\textbf{\text{Finish 1}}}$.

Solved with nukelauncher. We will show there exists an increasing sequence \((b_1,\ldots,b_n)\) satisfying \[\sum_{i\in X}b_i=\sum_{i\notin X}b_i.\]This clearly suffices by scaling. Assume without loss of generality that \(\sum_{i\in X}a_i<1\), since otherwise we replace \(X\) with \(\{1,\ldots,n\}\setminus X\). Then if \(n\in X\), we win automatically by letting \(b_i=a_i\) for \(i\le n-1\) and varying \(b_n\ge a_n\). Henceforth \(n\notin X\). Claim: We have \[\sum_{i\in X}a_{i+1}>2-\sum_{i\in X}a_i.\] Proof. Since \(\sum_{i\in X}a_{i+1}\) is greater than \(\sum_{i\in X}a_i\) but farther from 1 than \(\sum_{i\in X}a_i\), we have \(\sum_{i\in X}a_{i+1}\ge2-\sum_{i\in X}a_i\). Suppose that equality holds. Evidently there is some \(i\in X\) with \(i+1\notin X\); then replacing \(i\) with \(i+1\) in \(X\) yields a sum closer to 1 than \(\sum_{i\in X}a_i\), contradiction. \(\blacksquare\) Consider the sequence \((c_1,\ldots,c_n)\) defined by \(c_i=a_{i+1}\) for \(i\in X\) and \(c_i=a_i\) otherwise. Claim: We have \[\sum_{i\in X}c_i>\sum_{i\notin X}c_i.\] Proof. By the first claim, \[\sum_{i\in X}c_i=\sum_{i\in X}a_{i+1}>2-\sum_{i\in X}a_i=\sum_{i\notin X}a_i=\sum_{i\notin X}c_i,\]proving the claim. \(\blacksquare\) Finally, we have \[\sum_{i\in X}a_i<\sum_{i\notin X}a_i\quad\text{and}\quad\sum_{i\in X}c_i>\sum_{i\notin X}c_i.\]We are free to vary \(a_i\le b_i<c_i\) for \(i\in X\), so evidently there is an increasing sequence \((b_1,\ldots,b_n)\) satisfying \[\sum_{i\in X}b_i=\sum_{i\notin X}b_i,\]and we are done.

Is this solution correct? Note that since the $a_{i}$'s sum to $2$, set $\bar{X} = \{1, 2, \ldots, n\} - X$ also satisfies the problem's minimization condition. Therefore, WLOG we can assume that $\sum_{i \in X} a_{i} = 1 - \epsilon$, where $\epsilon > 0$. We consider two cases: Case 1: There is an index $j$, $1 \le j \le n -1$, such that $a_{j} \in X$ and $a_{j + 1} \in \bar{X}$. In this case, we have $a_{j + 1} - a_{j} \ge 2\epsilon$. Otherwise, adding $j + 1$ to $X$ and removing $j$ from it yields a new set $X ^ {\prime}$ closer to $1$ than $X$, contradicting the problem's minimization condition. Therefore, we can define $b_{j} = a_{j} + \epsilon$, $b_{j + 1} = a_{j + 1} - \epsilon$, and $b_{k} = a_{k}$ for every other $1 \le k \le n$. Case 2: $j > k$ for every $j \in X$ and $k \in \bar{X}$. Note that in this case $a_{1} \in \bar{X} \ge 2\epsilon$. Otherwise, adding $1$ to $X$ yields a new set $X ^ {\prime}$ closer to $1$ than $X$, contradicting the problem's minimization condition. Therefore, we can define $b_{n} = a_{n} + \epsilon$, $b_{1} = a_{1} - \epsilon$, and $b_{k} = a_{k}$ for every other $2 \le k \le n - 1$.

Solution. Let $|X|=p$. Let $X_1,...,X_p$ be the elements of $X$ in increasing order. WLOG $$a_{X_1}+...+a_{X_p}=1+s$$where $s>0$. The main idea is to show that $(a_{X_1}-s)+a_{X_2}+..+a_{X_p}=1$ works. We finish the problem by proving three things. 1. $a_{X_1}-s$ is positive. 2. We need to show that the order is still preserved: $a_{X_1}-s>a_{X_1-1}$. Note that if $X_1=1$, we're done. We assume that $X_1 \ge 2$. 3. To make the sum $2$, we show that there is an $i \notin X$, such that $a_i+s<a_{i+1}$ (Again, to preserve the order). Claim 1. $a_{X_1}-s$ is positive Proof. Because of the minimality condition, $$a_{X_2}+..+a_{X_p}=s+1-a_{X_1} \leq 1-s$$$$2s\leq a_{X_1} \implies a_{X_1}-s>0$$Claim 2. $a_{X_1}-s>a_{X_1-1}$ Proof. $$a_{X_2}+..+a_{X_p}+a_{X_1-1}=s+1-a_{X_1}+a_{X_1-1}\leq 1-s$$$$2s\leq a_{X_1}-a_{X_1-1} \implies a_{X_1}-s>a_{X_1-1}$$Claim 3. Point 3, above.. Proof. let $a_i$ be the largest value such that $i \notin X$. If $a_i=a_n$, then just add $a_i$ by $s$. Otherwise, $a_{i+1} \in X$. Similar to Claim 2, we have that $a_{i+1}-a_i \ge 2s \implies a_i+s < a_{i+1}$.

In my opinion not that hard for an A3, but the amount of intimidating notation makes solvng the problem and writing up the solution considerably harder.

Let $\sum_{i\in X} a_i = 1-\varepsilon$. WLOG $\varepsilon>0$, since the complement sums to $1+\varepsilon$. Lemma 1: Suppose $i \in X$ and $j \not \in X$, and $i<j$. Then $a_j-a_i \ge 2\varepsilon$. Proof: We know $a_i<a_j$ since the sequence is increasing. Suppose instead that $0<a_j-a_i< 2\varepsilon$. Then add $j$ to $X$ and remove $i$, so the new sum is $1-\varepsilon+a_j-a_i < 1+\varepsilon$. Now this new set is closer to $1$ than the original $X$ was, contradicting minimality. $\blacksquare$ Lemma 2: Suppose $j\not \in X$. Then $a_j \ge 2\varepsilon$. Proof: If $a_j < 2\varepsilon$, then adding $j$ to $X$ makes the new sum between in the range $(1-\varepsilon, 1+\varepsilon)$, contradicting minimality. $\blacksquare$ Suppose there exists an index $i$ for which $a_i \in X$ and $a_{i+1} \not \in X$. By Lemma 1, $a_{i+1} - a_i \ge 2\varepsilon$. Create a new sequence where we increase $a_i$ by $\varepsilon$ and decrease $a_{i+1}$ by $\varepsilon$. This change preserves the strictly increasing property and all positive property, so we have a new sequence in which the $X$-indexed elements sum to $(1-\varepsilon)+\varepsilon=1$, finishing. (Technically, the change could make $a_i$ and $a_{i+1}$ equal, so increase $a_i$ by $\delta \ll \varepsilon$ less than $\varepsilon$ and increase some other $X$-indexed element by $\delta$, where $\delta$ is small enough so that this preserves strictly increasing.) Else, we must have $X=\{k+1,\ldots,n\}$ for some $k\ge 1$. This guarantees that $a_1\not \in X$ and $a_n\in X$, so simply remove $\varepsilon$ from $a_1$ and add $\varepsilon$ onto $a_n$. This new sequence sums to $(1-\varepsilon)+\varepsilon=1$. It preserves all positive since $a_1 \ge 2\varepsilon>\varepsilon$ by Lemma 2, and it obviously preserves the strictly increasing property.

Note that either $S = \sum_{i\in X} a_i = 1 \pm \varepsilon,$ but since the sum is two let us WLOG consider $1-\varepsilon.$ Suppose for the sake of contradiction that no such sequence of $b_i$ exists. Claim: For some integer $1\le k\le n,$ $a_1, a_2,\dots, a_k\not\in X$ and $a_{k+1}, \dots, a_n \in X.$ Proof. Suppose not. Consider $X$ such that $T = \sum_{i\in X} i$ is maximal. Then there must be some $i,j$ with $i<j$ and $a_j\not\in X$ and $a_i\in X.$ Note that $a_j-a_i\ne 2\varepsilon$ otherwise we could replace $i$ with $j$ in $X,$ contradicting maximality of $T$. Moreover, if $a_j-a_i<2\varepsilon$ then replacing $i$ with $j$ in $X$ contradicts minimality of $S.$ As a result, $a_j-a_i > 2\varepsilon,$ so setting $b_i = a_i + \varepsilon$ and $b_j = a_j-\varepsilon$ and $b_k = a_k$ for $k\ne i,j$ gives a valid sequence of $b,$ contradiction. $\blacksquare$ Claim: For all $i\not\in X,$ $a_i \ge 2\varepsilon.$ Proof. If otherwise then add $i$ in $X,$ contradicting minimality. $\blacksquare$ Finally set, \begin{align*} b_i = a_i - \frac{\varepsilon}{k} &\qquad 1\le i\le k \\ b_i = a_i + \frac{\varepsilon}{n-k} &\qquad k+1\le i\le n, \end{align*}Whereby for $1\le i\le k,$ $b_i = a_i-\varepsilon/k \ge \varepsilon(2-1/k) > 0,$ giving a valid sequence for $b.$

I think I have quite a simple sketch of the solution but I have not worked it out on paper, so it might be a fake-solve. Sketch: We wish to prove the contrapositive of the statement. i.e. If the desired $\{ b_i \}$ does not exist then the first absolute sum is not minimised. Lemma: The set $X$ satisfies the second condition iff $X$ neither majorizes $Y$ or be majorized by $Z$, where $$Y=\{n, n-2, n-4, \dots ,2 \}, Z=\{n-1, n-3, n-5, \dots 1\} $$$$Y=\{n, n-2, n-4, \dots 1\}, Z=\{ n-1, n-3, n-5, \dots 2\}$$for even and odd $n$ respectively. Note that if $|A| > |B|$ then $B$ cannot majorize $A$, but $A$ can majorize $B$. If $X$ majorizes $Y$, then \[ \sum_{i \in X} a_{i} > \sum_{i \in Y} a_{i} > 1 \implies \left|1-\sum_{i \in X} a_{i}\right| > \left|1-\sum_{i \in Y} a_{i}\right| \]If $X$ is majorized by $Z$, then \[ \sum_{i \in X} a_{i} < \sum_{i \in Z} a_{i} < 1 \implies \left|1-\sum_{i \in X} a_{i}\right| > \left|1-\sum_{i \in Z} a_{i}\right| \]which will imply that the first sum is not minimised anyways. Contradiction.

The idea is to characterize all possible $X$. I am minimizing $$|\sum_{i\in [n]\backslash X} a_i - \sum_{i\in X} a_i|$$ If $X=\{x_1,\cdots,x_t\}$ and there exists $z_j\in [n]\backslash X$ such that $z_j>x_j$, then I should swap $(z_i,x_i)$ to reduce that absolute value. There is a similar condition for $[n]\backslash X$ (*) Now, our goal is to construct $c_1<c_2<\cdots<c_n, d_1<\cdots<d_n$ such that $$\sum\limits_{i\in [n]\backslash X} c_i - \sum\limits_{i\in X} c_i < 0 < \sum\limits_{i\in [n]\backslash X} d_i - \sum\limits_{i\in X} d_i $$ Then let $-A=\sum\limits_{i\in [n]\backslash X} c_i - \sum\limits_{i\in X} c_i$, $B=\sum\limits_{i\in [n]\backslash X} d_i - \sum\limits_{i\in X} d_i$, then setting $$b_1:b_2:\cdots:b_n = Bc_1+Ad_1 : Bc_2+Ad_2 : \cdots : Bc_n+Ad_n$$works WLOG $n\in [n]\backslash X$, then set $d_j=j$ for all $1\le j\le n-1$ and $d_n=10^{10^n}$ It remains to construct $c_i$. Let $X=\{x_1,\cdots,x_t\}, [n]\backslash X=\{y_1,\cdots,y_{n-t}\}$, and consider the smallest integer $j$ such that $x_{t-j}>y_{n-t-j}$. It must exist by (*), then consider the following construction: for all $k>x_{t-j}, c_k=10^{10^n}+k\epsilon^2 10^{-10^n}$ $x_{t-j}=10^{10^n}, y_{n-t-j}=10$ and for all $k<y_{n-t-j}$, $c_k=k\epsilon^2 10^{-10^n}$ Check this works.

Let the elements of $X$ be $c_1$, $c_2$, $c_3$, $\dots$, $c_i$. Without loss of generality, we can assume that \[a_{c_1}+a_{c_2}+a_{c_3}+\dots+a_{c_i}=1+d > 1\]If $d=0$ we are already done by making the sequence $a$ the same as the sequence $b$. Now, we will show that we can decrease a variable in $a_X$ by $s$, increase a variable not in $a_X$ by $s$, and maintain the relation $0<a_1<a_2<\dots < a_{n-1}<a_n$. Let us decrease $a_{c_j}$ by $d$, then we will show that $a_{c_j} - d > a_{c_j-1}$. Clearly, \[a_{c_1}+\dots+a_{c_{j-1}}+a_{c_j-1}+a_{c_{j+1}}+\dots+c_i\ge 1-d\]due to minimality. Now, we have $a_{c_j}-a_{c_j-1}\ge 2d$ which means we can subtract $d$ from $a_{c_j}$ without getting into any trouble. The same applies to adding $d$ to a non-$a_X$.

Idea is somewhat similar to 2022 USAJMO 2. Essentially, if there exist sequences $(b_1,b_2,\dots,b_n)$ and $(c_1,c_2,\dots,c_n)$ where the second condition has the first sum less and the second sum greater, some sort of smoothing argument that changes $b_i$ to $c_i$ will give a correct value. Then $X$ must be majorized by its complement (WLOG) and we can find some better subset than $X$ by replacing an element of $X$ with the corresponding element in $X^C$. Done.

Hopefully, it isn't fakesolve. WLOG, we can assume $\sum_{i \in X} a_i < 1$. Let $\varepsilon = 1 - \sum_{i \in X} a_i$.Let $i_1 < i_2 < \dots < i_k$ be the elements of $X$. We split the problem into two cases: Case 1: $n \in X$. If there exists $s$ such that $i_{s + 1} > i_s + 1$, then $i_s + 1 \not\in X$, which means that $\sum_{j \in X \backslash \{i_s\} \cup \{i_s + 1\}} a_j \ge 1 + \varepsilon$, thus $a_{i_s + 1} \ge a_{i_s} + 2\varepsilon$, so replacing $a_n \to a_n + \varepsilon$ and $a_{i_s + 1} \to a_{i_s + 1} - \varepsilon$ works. If there doesn't exists such $s$, then $X = \{n, n - 1, n - 2, \dots, n - k + 1 \}$. Then we can easily scale it. Case 2: $n \not\in X$. Let $m = i_k$. Then considering $X \backslash \{m\} \cup \{m+1\}$, we get $a_{m+1} - a_m \ge 2\varepsilon$. If $a_{m+1} > a_m + 2\varepsilon$, then replacing $a_{m+1} \to a_{m+1} - \varepsilon$ and $a_m \to a_m + \varepsilon$ works. Thus we may assume $a_{m+1} - a_m = 2\varepsilon$. If there exists $k \neq j \in X$, then replace $a_{m+1} \to a_{m+1} - \varepsilon$ and $a_m \to a_m + \varepsilon_1$ and $a_j \to a_j + \varepsilon - \varepsilon_1$ and take $\varepsilon_1$ such that $a_{j+1} > a_j + \varepsilon - \varepsilon_1$. Otherwise, since $n \ge 3$, there exists $j$ different from $k, k + 1$ and $j \not\in X$. Then replace $a_m \to a_m + \epsilon$ and $a_{m+1} \to a_{m+1} - \varepsilon_1$ and $a_j \to a_j - (\varepsilon - \varepsilon_1)$ and take $\varepsilon_1$ such that $\varepsilon - \varepsilon_1$ is close to $0$. Then it works perfectly. Hence, in all cases, we can scale the sequence $(a_i)_{1 \le i \le n}$, so we're done. $\blacksquare$

Obviously redefine $X$ to be a subset of $\{a_1,\dots,a_n\}$. Let $s=\displaystyle{\sum_{i\in X}i}<1$ and let $Y$ be the complement of $X$. Suppose there exist $y\in Y$ and $x\in X$ where $y>x$. Then consider set $X'$ formed by removing $x$ and adding $y$. Clearly \[s-x+y\ge 2-s\implies y-x\ge 2-2s.\]Now if there exist $y\in Y$ and $x\in X$ where $y>x$ at all, then there must exist $i$ such that $a_{i+1}\in Y$ and $a_i\in X$. Consider the sequence $b$ where If $j\not\in \{i,i+1\}$ then $b_j:=a_j$. We have $b_i:=a_i+(1-s)$ and $b_{i+1}:=a_{i+1}-(1-s)$. This works provided that $a_{i+1}-a_i\ne 2-2s$. In this case set $b_i=a_i+(1-s)-\epsilon$ instead and increase some other element in $X$ by $\epsilon$. Now suppose that there do not exist $y\in Y$ and $x\in X$ such that $y>x$. In this case divide all values by $2-s$ so that the sum of values in $Y$ is $1$. Now increase the largest value in $X$ by \[1-\frac{s}{2-s}=\frac{2-2s}{2-s}\]and we are done. $\blacksquare$

Giving explicit constructions is too hard We will characterize all sets $X$ that could possibly satisfy the given condition, then show that such a sequence $\{b_i\}$ exists for all such sets $X$. We say that a set $X$ dominates a set $Y$ if we can assign a distinct $f(y) \in X$ for each $y \in Y$ such that $f(y) > y$ for all $y$. Claim. [First Part] For any sequence $\{a_i\}$, the given value can only be minimized when neither $X$ nor $X^c$ dominate each other. Proof. Assume otherwise, and suppose $X^c$ dominates $X$. Let $\sum_{i \in X} a_i = 1-\varepsilon$ for $\varepsilon > 0$. If $|X| = 1$, assume for the sake of contradiction that $i < n$. It follows that $1-\varepsilon = a_i < a_n \leq 1+\varepsilon$, so $a_n$ is closer to $1$ than $a_i$, contradiction. Suppose that $|X| \geq 2$. It follows that for every index $i \in S$, the difference $a_{f(i)} - a_i$ is at least $2\varepsilon$, as otherwise, by replacing $i$ with $f(i)$ in $X$ we get a sum closer to $1$. Hence \[\sum_{j \in X^c} a_j - \sum_{i \in X} a_i \geq 2\varepsilon |X| > 2\varepsilon\]yields a contradiction. $\blacksquare$ Claim. [Second Part] If neither $X$ nor $X^c$ dominates the other, then we can find such a sequence $\{b_i\}$. Proof. Let $|X| = r$. It suffices to construct a sequence $\{b_i\}$ such that $\sum_{i \in X} b_i = \sum_{j \in X^c} b_j$, as we can scale by a suitable amount. Thus, we will instead construct a sequence $(c_1, c_2, \dots, c_n)$ with $c_i > 0$ for each $i$ such that \[\sum_{i=0}^{r-1} i \sum_{j=x_i}^{x_{i+1}} c_j = \sum_{i=0}^{r-1} (n-r-i) \sum_{j = y_i}^{y_{i+1}} c_j\]where $\{x_i\}$ is the increasing sequence of all indices in $X$ and $\{y_i\}$ is the increasing sequence of all indices in $X^c$. Then, taking $b_i = c_1+c_2+\cdots+c_i$ for each $i$ does the trick. (Here $c_i$ is just the difference sequence of $b_i$.) Notice that every $c_i$ appears $f(i)$ times in the LHS sum, where $f(i)$ denotes the number of elements in $X$ at least $a_i$. This element is then counted $n-i+1 - f(i)$ times in the RHS sum, as it is counted $n-i+1$ times total. So it suffices to demonstrate such a sequence $\{c_i\}$ with $c_i > 0$ satisfying only the condition \[\sum_{i=1}^n c_i(2f(i) - n+i-1) = 0.\]There obviously exists such a $\{c_i\}$ as long as $2f(i) - n+i-1 > 0$ and $2f(j) - n+j-1 < 0$ for some $i, j$. Assume for the sake of contradiction that $2f(i) \leq n-i+1$ for all $i$; the other case is analogous, say by swapping $X$ with $X^c$. We will arrive at a contradiction by inductively pairing off every index $x_i \in X$ with a $y_j \in X^c$ with $x_i \leq y_j$. For our base case, notice that for $i=n$, the inequality reads $2f(n) \leq 1$, i.e. $f(n) = 0$, so $n \not \in X$. So we can pair $x_r$ with $y_{n-r} = n$. For the inductive step, for $i = x_j$ the inequality reads $f\left(x_j\right) \leq \frac{n-x_j + 1}2$. In particular, there are at least $\frac{n-x_j + 1}2$ elements $y_i$ among $x_j + 1, \dots, n$. Because there are only at most $\frac{n-x_j-1}2$ elements $x_i$ among these which we have paired off with an equal number of $y_j$, there exists at least $\frac{n-x_j+1}2 - \frac{n-x_j-1}2 = 1$ unpaired $y_\ell$, which we can pair with our $x_j$, completing the induction. Thus we can create a pairing $(x_i, y_j)$ for all $x_i \in X$, which contradicts the assumption. It follows that such a sequence $\{c_i\}$ must exist, completing the problem.

Why do everyone's solution include majorization? I find the construction the same as @goodar2006, which I think is quite intuitive. Call $X$ good if there is $a_n$ satisfying the constraint so $ | \sum_{i \in X} a_i - 1 | $ is minimised. We wish to modify $a_n$ while keeping its stricty monoticity and sum, so that $\sum_{i \in X} a_i=1$. Since $X$ is good if and only if its complement $\bar{X}$ is, we can assume without loss of generality that $\sum_{i \in X} a_i < 1$.

The rigorous way of writing the solution follows: Pick $k$ so $k \in X$ but $k \notin X$, we claim $a_k + (1-s) < a_{k+1} - (1-s)$, where $s = \sum_{i \in X} a_i$. Indeed, let $X' = X + \{k+1\} - \{k\}$, then by minimality we have $|s - a_k + a_{k+1} - 1| \geq |s - 1| = 1-s$. This simplifies to $a_{k+1}-(1-s) \geq a_k+(1-s)$. So we can let $b_k = a_k+1-s$, $b_{k+1} = a_{k+1}-(1-s)$ and $b_m=a_m$ otherwise. This keeps the monoticity except when equality is a held, but in this case we can decrease $a_k$ by some $\epsilon$ and then add to any other $a_i$ with $i \in X$. This is unless $X$ only has one element (which is why $n = 2$ is excluded). If this is the case, we have two cases: 1. $X = \{1\}$. Then $2 - (a_3 + ...) = a_1 + a_2 > 2 - a_1 \implies a_3 \leq a_3+...<a_1$, a contradiction. 2. $X = \{m\}$ for some $m \geq 2$. If $a_m \leq \frac{1}{2}$, then $a_1 + a_m < 1$ which is closer to $1$ than $a_m$, contradiction. If $a_m > \frac{1}{2}$, then $a_1 + a_m > 2 - a_m \implies 2-(a_1+a_m) < a_m$. L.H.S. is the sum of all $a_i$ except $i=1,m$, so this it must be the case that $m = n$. In this case, we can still indeed construct $b_i$ so $b_n = 1$ while keeping $0<b_1 < b_2 < ... < b_n$. If we cannot pick such $k$ describe above, then $X = \{m, m+1, ..., n\}$ for some $m$ (X = $\phi$ is not possible). By using the minimality on the set $X = X + \{ a_1 \}$ we have that $a_1 - (1-s) \geq 1-s > 0$, so we can subtract $1-s$ from $a_1$ and add to $a_n$, keeping the positivity, strict monoticity and giving the desired sum.

goodar2006 wrote: Is this solution correct? Note that since the $a_{i}$'s sum to $2$, set $\bar{X} = \{1, 2, \ldots, n\} - X$ also satisfies the problem's minimization condition. Therefore, WLOG we can assume that $\sum_{i \in X} a_{i} = 1 - \epsilon$, where $\epsilon > 0$. We consider two cases: Case 1: There is an index $j$, $1 \le j \le n -1$, such that $a_{j} \in X$ and $a_{j + 1} \in \bar{X}$. In this case, we have $a_{j + 1} - a_{j} \ge 2\epsilon$. Otherwise, adding $j + 1$ to $X$ and removing $j$ from it yields a new set $X ^ {\prime}$ closer to $1$ than $X$, contradicting the problem's minimization condition. Therefore, we can define $b_{j} = a_{j} + \epsilon$, $b_{j + 1} = a_{j + 1} - \epsilon$, and $b_{k} = a_{k}$ for every other $1 \le k \le n$. Case 2: $j > k$ for every $j \in X$ and $k \in \bar{X}$. Note that in this case $a_{1} \in \bar{X} \ge 2\epsilon$. Otherwise, adding $1$ to $X$ yields a new set $X ^ {\prime}$ closer to $1$ than $X$, contradicting the problem's minimization condition. Therefore, we can define $b_{n} = a_{n} + \epsilon$, $b_{1} = a_{1} - \epsilon$, and $b_{k} = a_{k}$ for every other $2 \le k \le n - 1$. I think this is almost correct except when an equality case occurs (in your case 1, "closer to $1$ than $X$" is almost true, but it could be the case that they are both equidistant).