If $n=3$ use the set $\{1,2,3,4,5,7\}$. For $n>3$ construct $$S = \{F_1,F_2,F_3,...,F_{2n-1}\}\cup\{F_{2n-2}-2\}$$and partition \begin{align*} A_m &= \{F_{2n-2m+3},F_{2n-2}-2\}\cup\{F_{2n-2m+4},F_{2n-2m+6},F_{2n-2m+8},...,F_{2n-2}\} \\ B_m &= S \setminus\ A_m. \end{align*}where $F_n$ denote $n^{th}$ Fibonacci number. Seeing that $|A_m| = m$ then a straightforward calculation shows that this partition works.

Here is a construction for $n = 5$ which generalizes readily: we write \begin{align*} S &= \big\{ 1, 11, 111, 1111, 11111, \\ &\qquad 12, 123, 1234, a, b \big\} \end{align*}where $a$ and $b$ are integers chosen so that the first five elements have sum equal to the last five elements. Observe that the two halves have equal sum. Then we have \begin{align*} 1 + 11 + 111 + 1111 + 11111 &= 12 + 123 + 1234 + a + b \\ 12 + 111 + 1111 + 11111 &= 1 + 11 + 123 + 1234 + a + b \\ 123 + 1111 + 11111 &= 1 + 11 + 111 + 12 + 1234 + a + b \\ 1234 + 11111 &= 1 + 11 + 111 + 1111 + 12 + 123 + a + b \end{align*}which gives the desired equalities for $m = 2, 3, 4, 5$.

In my country's TST camp. There are many different solutions(at least 3).

I guess I will share my construction as well. If $n=3$ then just take $\{1,2,3,4,5,7\}$. If $n>3$, take $$S = \{2^1,2^2,...,2^{n-1}\}\cup\{2^1-1,2^2-1,...,2^n-1\}\cup\{2^n-n-2\}.$$and partition \begin{align*} A_m &= \{2^n-n-2\}\cup\{2^{n-1},2^{n-2},2^{n-3},...,2^{n-m+2}\}\cup\{2^{n-m+2}-1\} \\ B_m &= \{2^1,2^2,2^3,...,2^{n-m+1}\}\cup \{2^1-1,2^2-1,...,2^n-1\}\setminus\{2^{n-m+2}-1\}. \end{align*}A straightforward calculation shows that this works.

psi241 wrote: Let $n\geqslant 3$ be an integer. Prove that there exists a set $S$ of $2n$ positive integers satisfying the following property: For every $m=2,3,...,n$ the set $S$ can be partitioned into two subsets with equal sums of elements, with one of subsets of cardinality $m$. https://artofproblemsolving.com/community/c6h455908p2561601 Not such good problem

We do $n=3,4$ by giving examples: \begin{align*} n=3: & \quad \quad S=\{2,3,4,6,7,8\} \\ n=4: & \quad \quad S=\{1,2,3,4,7,8,10,15\} \end{align*}and the relevant sets are: \begin{align*} n=3: & \quad \quad \{7,8\} \, , \, \{3,4,8\} \\ n=4: & \quad \quad \{10,15\} \, , \, \{2,8,15\} \, , \, \{1,2,7,15\} \end{align*}We now go by induction on $n$. Let $S_n$ be the working set for $n$ and let $2T$ be the total of all the elements in $S_n$. Then define $A_k \subset S_n$ to be the subset of size $2 \leq k \leq n$ with total $T$. We define: $$S_{n+2}=S_n \cup \{T,2T,3T,4T\}$$ This has $2n+4$ elements as $S_n$ cannot contain an element $\geq T$ by considering $A_2$ and $S_n \setminus A_2$ which both have total of elements $T$ and neither can contain an element $\geq T$. Also the total of the elements in $S_{n+2}$ is $10T+2T=12T$. Now we can define $B_k \subset S_{n+2}$ as follows: \begin{align*} k=2: & \quad \quad B_2=\{2T,4T\} \\ k=3: & \quad \quad B_3=\{T,2T,3T\} \\ k \geq 4: & \quad \quad B_k=\{3T,2T\} \cup A_{k-2} \end{align*}and all the $B_k$ have total $6T$. As we've covered $n=3,4$ and shown we can go $n \rightarrow n+2$ by induction we're done for all $n \geq 3$

Do the integers in $S$ have to be distinct? It doesn't say in the problem, but if they don't have to be, then it is solvable by a simple induction. What am I missing?

JANMATH111 wrote: Do the integers in $S$ have to be distinct? It doesn't say in the problem, but if they don't have to be, then it is solvable by a simple induction. What am I missing? The problem asks for a set of positive integers, which implicitly means that all $2n$ elements must be distinct.

MarkBcc168 wrote: I guess I will share my construction as well. If $n=3$ then just take $\{1,2,3,4,5,7\}$. If $n>3$, take $$S = \{2^1,2^2,...,2^n-1\}\cup\{2^1-1,2^2-1,...,2^n-1\}\cup\{2^n-n-2\}.$$and partition \begin{align*} A_m &= \{2^n-n-2\}\cup\{2^{n-1},2^{n-2},2^{n-3},...,2^{n-m+2}\}\cup\{2^{n-m+2}-1\} \\ B_m &= \{2^1,2^2,2^3,...,2^{n-m+1}\}\cup \{2^1-1,2^2-1,...,2^n-1\}\setminus\{2^{n-m+2}-1\}. \end{align*}A straightforward calculation shows that this works. Your set has only $2n-1$ elements, it can easily be corrected by replacing $\{2^{n}-n-2\}$ by $\{2^{n}-n-3,1\}$

There was a typo in the original post. Thanks for noticing.

I might be wrong, but there's actually no need for a construction...? *Warning* The following is just a SKETCH and could be incorrect: Let $S=\{ a_1,a_2,\cdots,a_n\}$. WLOG let $a_1+a_2+\cdots+a_{2n}=1$ (because we can cancel out the denominator to turn rationals into integers) then we only need the following system of equations to hold: \begin{align*} a_{2n}+a_{2n-1}&=\frac12,\\ a_{2n}+a_{2n-2}+a_{2n-3}&=\frac12,\\ a_{2n}+a_{2n-3}+a_{2n-4}+a_{2n-5}&=\frac12,\\ \cdots\\ a_{2n}+a_{n}+a_{n-1}+\cdots+a_1&=\frac12. \end{align*}From this we can obtain $a_{2n-k}=a_{2n-2k}+a_{2n-2k-1}$. Note that there's only one constraint regarding $a_1,a_2,\cdots,a_n$: Sum up $a_{2n-k}=a_{2n-2k}+a_{2n-2k-1}$ for $k=1,2,\cdots,n-1$ to get $a_1+a_2+\cdots+a_n=a_{2n}$. Then the values of $a_1,a_2,\cdots,a_n$ can determine $a_1,a_2,\cdots,a_{n+\lceil \frac{n}{2}\rceil}$. Repeat the above reasoning for multiple times, and you'll determine the value of $a_1,a_2,\cdots,a_{2n}$. In the meanwhile we can allow $a_{2n}$ to be considerably small, so that $a_1+a_2+\cdots+a_{2n}=1$ still holds. $\square$

I will present another construction. Define sequence $(a_i)_{i=1}^{\infty}$ as follows: $a_1=1$, $a_2=2$, $a_3=3$. $$a_{4k}=a_{4k-2}+a_{4k-1}$$$$a_{4k+1}=a_{4k-1}+a_{4k-2}+a_{4k-3}$$$$a_{4k+2}=a_{4k+1}+1$$$$a_{4k+3}=a_{4k+2}+1$$For example, our sequence starts $1, 2, 3, 5, 6, 7, 8, 15, 21, 22, 23, 45, 66, 67, 68, 135$, etc. Case I: $n=2x+1$. Then let $$S=\{a_1, a_2, \cdots, a_{4x}, a_{4x+1}, \sum_{i=1}^{4x} a_i - a_{4x+1}\}$$It is clear that $a_1 < a_2 < \cdots < a_{4x} < a_{4x+1} < \sum_{i=1}^{4x} a_i - a_{4x+1}$. Now let $A=\sum_{i=1}^{4x} a_i - a_{4x+1}$. $$a_{4x+1}+A=\frac{1}{2}\sum_{s\in S} s$$Also, note that $$a_{4k+1}+a_{4k+2}+a_{4k+3}=a_{4k+1}+a_{4k+4}=a_{4k+5}$$Hence, \begin{align*} a_{4x+1}+A&=a_{4x-3}+a_{4x}+A\\ &= a_{4x-3}+a_{4x-2}+a_{4x-1}+A\\ &=a_{4x-7}+a_{4x-4}+a_{4x-2}+a_{4x-1}+A\\ &=a_{4x-7}+a_{4x-6}+a_{4x-5}+a_{4x-2}+a_{4x-1}+A\\ &=\cdots \end{align*}Hence we have provided a construction for $n$ odd. Case II: $n=2x$ Then use a similar argument so that $$S=\{a_1,a_2,\cdots,a_{4x-2},a_{4x-3}+a_{4x-2},\sum_{i=1}^{4x-4}a_i\}$$It is clear that $$a_1 < a_2 < \cdots < a_{4x-3} < a_{4x-2} < a_{4x-3}+a_{4x-2} < \sum_{i=1}^{4x-4} a_i$$Using a similar argument to the odd case, the set $S$ also satisfies the condition. Hence we have provided a construction for $n$ even.

Nice problem, but indeed very easy Let $\mathcal{T}_m$ be the subset of cardinality $m$ satisfying the problem condition. Now, consider the sequence $t_k$ of positive integers such that $$t_{2k+1}=t_{2k}+t_{2k-1},$$for all $k=2,3,\ldots,n-1$. Take $t_{2n}=t_1+t_2+\cdots+t_{2(n-2)}$. Now, choose the sets $\mathcal{T}_m$ to be $$\mathcal{T}_m = \{t_{2n},t_{2n-2},\ldots,t_{2n-2m+4}\}.$$Now, the only condition to be checked of is that $t_{2n}\neq \{t_1,t_2,\ldots,t_{2n-1}\}$ which is a pretty easy task!

Let the sequence $F_n$ be defined with $F_1=1, F_2=2,$ and $F_n = F_{n-1}+F_{n-2}$ for all $n \geq 3$. Also let $a_n = F_1+F_2+ \dots +F_{2n-4}$. I claim that the set $S = \{ F_1, F_2, \dots , F_{2n-1}, a_n \}$ works. First we show that the elements of this set are distinct; it suffices to show that $F_{2n-3} < a_n < F_{2n-2}$ for all $n \geq 3$. The lower bound is trivial; for the upper bound, we proceed by induction on $n$. The base case is easy; assuming that the inequality is true for $n = k$, then we note that $a_{k+1} = a_k+F_{2k-3}+F_{2k-2} < F_{2k}$ by the induction hypothesis, as desired. Now we show that we can indeed create the desired partitions of $S$. For every $m \in \{ 2, 3, \dots , n \}$ take the set $\{a_n, F_{2n-2}, F_{2n-4}, \dots , F_{2n-2m+4}, F_{2n-2m+3} \}$. It is easy to see that this set has cardinality $m$ and the sum of its elements is $a_n+F_{2n-1}$, which is half the sum of the elements of $S$. Thus this construction works and we're done.

Motivated by the following idea: If $a,b,a+b$ are in $S,$ then we can swap $a,b$ for $a+b$. For $n=3,$ just take $\{1,2,3,4,5,7\}.$ For $n>3$ the set $S$ is $\{F_1,F_2,\ldots,F_{2n-2},F_{2n-1},F_{1}+F_{2}+\ldots+F_{2n-4}\}.$ Let $F_1+F_2+\ldots+F_{2n-4}=X.$ Construction is $\{X,F_{2(n-1)},F_{2(n-2)},F_{2(n-3)},\ldots,F_{2(n-m+1)},F_{2(n-m)+1}\}.$

Just for the record. For $n=3$ $\{11,2^{2}+1\}\cup\{2^1+1,2^1,10,1\}$ For $n>3$ $\{\sum_{k=1}^{n-3}( 2^{k}+1+2^{k})+11,2^{(n-1)}+1\}\cup\{2^{n-2}+1,2^{n-2},...,2^1+1,2^1,10,1\}$ works for $n\geq 3$ in $n-2$ steps if in the $k$ step we move $2^{(n-k)}+1$ from the left subset to the right subset, and move $2^{(n-k-1)}+1 , 2^{(n-k-1)}$ from the right subset to the left subset.

It can also be done by induction. If we allow repeating integers (i think it is allowed), just take your set for k, and add the semisum of all its elements twice. You now have a set for k+1.

@above I don't think repeating elements is allowed rip since it's a set The fibbonaci construction seems much nicer, but here's another one (along the same lines): let $m = \frac{n(n-1)}{2}$, s = $1 + 2 + 3 + \dots + n-1 + (n-1 + n-2) + (n-1 + n-2 + n-3) + \dots + m$ and consider the set $$\left (1, 2, 3, \dots, n-1, n-1 + n-2, \dots, \frac{n(n-1)}{2}, \frac{n(n-1)}{2} + 1, 100^n, 100^n + s - 2m\right )$$and number the elements $(a_1, a_2, \dots, a_2n)$ in order. Then clearly $a_n = a_{n-1} + a_{n-2}$, $a_{n+1} = a_{n-1} + a_{n-2} + a_{n-3}$, etc. So we can take $(a_1, a_2, \dots, a_{n-1}, a_{2n}), (a_1, a_2, \dots, a_{n-3}, a_n, a_{2n}), (a_1, a_2, \dots, a_{n-4}, a_{n+1}, a_{2n}), \dots, (a_{2n-3}, a_{2n})$ as subsets of cardinality $n, n-1, \cdots, 3, 2$ respectively whose sum is half the sum of the big set.

Ok so here is a solution sketch. For $n = 3$, we just have ${1,2,3,4,5,7}$. Now, if $n$ works with a construction where the partition into subsets of $n$ elements each, with the two elements going to different sets (clearly true for $n = 3$), we prove a similar construction exists for $n+1$ To do this, if the two elements have difference $k$, we introduce two new large elements with difference $2k$. So, now, we just swap elements For example, to extend $n = 3$ into $n = 4$, we just add in $9$ and $13$, and our last partition becomes ${1,3,5,13}$ and ${2, 4, 7, 9}$. The construction for $m = 2$ becomes ${9, 13}$ and ${1, 2, 3, 4, 5, 7}$.

I refer to the set with cardinality $=m$ as the left set, and the one with cardinality $=2n-m$ as the right set. We use induction such that the largest element of the set is always in the left set. For the base case of $n=3$, we choose $\left\{1,2,3,4,5,7\right\}$. Check that $\left\{4,7\right\}=\left\{1,2,3,5\right\}$ and $\left\{1,3,7\right\}=\left\{2,4,5\right\}$ both have equal sums. Now we assume our induction hypothesis for some $n$ and we prove it for the $n+1$ case. For the $n+1$ case, notice that if we add some $x_1+x_2$ to the value of the largest element (which is always in the left set), and add the two $x_1$ and $x_2$ to the right, then we are actually done for all $m\in\left\{2,3,\ldots,n\right\}$. Thus we need to just handle the $m=n+1$ case manually. Also note that upon multiplying all the elements of our current set by some number, we still have a valid and working set. So we multiply all the elements of the $2n$ lengthed set by $2$. We are work with the $2n$ lengthed set from now on. Firstly, we pick $2$ elements say $l_1$ and $l_2$ from the left set that are distinct form the largest element. We also select another element from the right set say $r_1$. Now we take the construction where the left set and the right set both have $n$ elements. We add the numbers $x_1$ and $x_2$ to the right set and increase the value of the largest element in the left set by $x_1+x_2$. This is a working construction and the left set now has cardinality $=n$ and the right one has cardinality $=n+2$. And for the values of $x_1$ and $x_2$, we choose them in a way such that $x_1+x_2=l_1+l_2-r_1$. We first show that these values work, and then we show that it is possible to take such values. Firstly, swap $\left\{x_1,x_2,r_1\right\}$ with $\left\{l_1,l_2\right\}$. After this swapping, we have $n+1$ elements in the left set and $n+1$ elements in the right set. Easy checking suffices that the sum of both the sets are equal. Now for the part of showing we can actually pick such $x_1$ and $x_2$, notice that since we multiplied all the elements with $2$, we have that $x_1+x_2=l_1+l_2-r_1=\text{even}$, and so we pick $x_1$ and $x_2$ to be both odds and we are done for suitable choices of $l_1$, $l_2$ and $r_1$. P.S.: Aah, I noticed that there's actually a little flaw in the proof, but there's actually a simple fix to this. Notice that we can't take such $x_1$ and $x_2$ if we have that for all $r_1\ge l_1 + l_2>l_1$. For this we take the construction for the $m=n$ set, and then we can choose the $l_1$ as the least element of the left set. We then move $l_1$ to the right set and add $x_1$ and $x_2$ to the left set and $x_1+x_2$ to the value of the largest element of the left set. Also, we set $x_1+x_2=l_1$ which solves our problem. We can do this as we took $r_1>l_1$, so no element in the right set is smaller than $l_1$ and both $x_1$ and $x_2$ are smaller than $l_1$, so we are done.

Simply consider the set $S={f_1,f_2,\dots,f_{2n}}$ where $f_i$ is the $i^{th}$ number in the recursion; $$f_1 =1 \ \ \ f_2 = 2\ \ \ f_{n+2}=f_{n+1}+f_n $$This is just the Fibonacci Sequence starting from 1 and 2. Now, notice that we can have \begin{align*} \{f_1,f_2,\dots,f_{2n-2}\} &\text{ and } \{f_{2n-1},f_{2n}\}\\ \{f_1,f_2,\dots,f_{2n-4},f_{2n-1}\} &\text{ and } \{f_{2n-3},f_{2n-2},f_{2n}\}\\ &\vdots\\ \{f_1,f_2,f_5\dots,f_{2n-1}\} &\text{ and } \{f_3,f_4,\dots,f_{2n}\} \end{align*}Thus, for all $n\geq 3$, there exists a set $S$ of $2n$ distinct positive integers such that for any $m \in \{2,\dots,n\}$ the set $S$ can be partitioned into two sets with cardinalities $m$ and $2n-m$ with equal sums.

We can loosen the condition to allow positive rationals in $ S $ (just scale everything up by LCM of denominators to recover the desired set of positive integers). Consider the set \[ S = \left\{ \frac{9}{10}, \frac{9}{10^2}, \cdots, \frac{9}{10^{n-1}}, \frac{1}{10}, \frac{1}{10^2}, \cdots, \frac{1}{10^{n-1}}, x, y \right\} \]where $ x $ and $ y $ are arbitrary rationals such that the sum of all elements in $ S $ is $ 2 $. This is always possible as $ \frac{9}{10} + \frac{9}{10^2} + \cdots + \frac{9}{10^{n-1}} + \frac{1}{10} + \frac{1}{10^2} + \cdots + \frac{1}{10^{n-1}} < 1 + \frac{1}{9}$. Now for each $ m $ we pick one of the subsets to be \[ A = \left\{ \frac{9}{10}, \frac{9}{10^2}, ..., \frac{9}{10^{m-1}}, \frac{1}{10^{m-1}} \right\} \]It is easy to see that the sum of the elements in $ A $ is $ 1 $, so the remaining elements in $ S $ must sum to $ 2 - 1 = 1 $ too. (Just realized this is almost the same construction as #29)

Consider the sequence $f_1, f_2, f_3, \ldots$ where $f_1=1, f_2=2$, and $f_n=f_{n-1}+f_{n-2}$ for all $n\geq 3$. Now, we will prove that the set \[S=\left\{f_1, f_2,\ldots, f_{2n-1}, c\right\} \text{with } c=\sum_{i=1}^{2n-4}f_i\]satisfies our desired property. We start with the sets $A=\{f_1, f_2,\ldots, f_{2n-2}\}$ and $B=\{f_{2n-1}, c\}$. \[\sum_{i=1}^{2n-2}f_i = f_{2n-2}+f_{2n-3}+\sum_{i=1}^{2n-4}f_i=f_{2n-1}+c\]so $m=2$ works. Now, we'll show that we can keep on replacing an element of $B$ with two elements of $A$ which have the same sum, to reach all desired values of $m$. To do that, we'll use induction, showing that for every $m\in\{2,\ldots, n-1\}$, $B$ contains $f_{2n-(2m-3)}$ and $A$ contains $f_{2n-(2m-2)}$ and $f_{2n-(2m-1)}$. For $m=2$, we see that this is true, because $f_{2n-1}\in B$ and $f_{2n-2}, f_{2n-3}\in A$. Now, assume that it's true for $m=k$. Then, since $f_{2n-(2k-3)}=f_{2n-(2k-2)}+f_{2n-(2k-1)}$, we can swap $f_{2n-(2k-3)}$ with $f_{2n-(2k-2)}$ and $f_{2n-(2k-1)}$ to obtain the case where $m=k+1$, and we see that $B$ now contains $f_{2n-(2k-1)} = f_{2n-(2(k+1)-3)}$ and $f_{2n-(2(k+1)-2)}=f_{2n-2k}$ and $f_{2n-(2(k+1)-1)}=f_{2n-(2k+1)}$ are still in $A$ as long as $2k+1\leq 2n-1$, which simplifies to $k\leq n-1$. Hence, our induction is complete, and we're done.

Let the set of the first $2n-1$ Fibonacci numbers be $S=\{F_1=1, F_2=2, F_3=3, F_4=5, \ldots F_{2n-1}\}$. Let the sum of these be $F$. Then consider the set $S'=S \cup \{F-2F_{2n-1}=F_{2n-2}-2\}$. We can check this equality by a simple induction. Then $S'$ works. Basically the first set that works is ${F-2F_{2n-1},F_{2n-1}}$. After this, we can repeatedly replace the Fibonacci number in our partition with the two previous Fibonacci numbers, so it still adds to the same thing. The only other thing I have to do is check $n=3$, because in this case $F_{2n-2}-2$ is already a Fibonacci number and in our set. For this we can use the construction $\{1,4,5,11,16,6\}$.

We claim $S = \{F_1, F_2, \dots, F_{2n-1}, F_{2n-2} - 2\}$, where $F_i$ is the $i+1$th Fibonacci number works. Note that $\{F_{2n-1}, F_{2n-2} - 2\}, \{F_{2n-3},F_{2n-2}, F_{2n-2} - 2\}, \dots ,\{F_1, F_2, F_4, \dots, F_{2n-2}, F_{2n-2} -2\}$ work since $F_{2n-2} - 2 = F_1 + F_2 + \dots + F_{2n-4}$.

If $n = 3$, then we can let $S = \{1, 2, 3, 5, 6, 7\}$ whereupon we can choose the subsets \[\{5, 7\}, \{1, 2, 3, 6\}\]and \[\{1, 5, 6\}, \{2, 3, 7\}\]for $m = 2$ and $3$ respectively. Else let $S$ be \[\left\{3^{n - 1}, \frac{3^{n - 1} - 3}{2}\right\} \cup \bigcup_{0 \leq i \leq n - 2} \{3^i, 2 \cdot 3^i\}\text{.}\]This is a well-defined set since \begin{align*} \frac{3^{n - 1} - 3}{2} &= \frac{3(3^{n - 2} - 1)}{2} \\ &= 3(1 + 3 + \dots + 3^{n - 3}) \\ &= 3 + 3^2 + \dots + 3^{n - 2} \end{align*}and since $n - 2 \geq 2$, this sum has at least two powers of $3$ and is thus not equal to any other element in $S$ by comparing their base-$3$ representations. Now we show that this choice of $S$ actually works. This argument is clearer if we let $n$ be, say, $5$: We start with the two subsets \[\left\{3^4, \frac{3^4 - 3}{2}\right\}, \{1, 2, 3, 2 \cdot 3, 3^2, 2 \cdot 3^2, 3^3, 2 \cdot 3^3\}\]which evidently work. Now we swap the $3^4$ on the left-hand side with the $3^3$ and $2 \cdot 3^3$ on the right-hand side: \[\left\{3^3, 2 \cdot 3^3, \frac{3^4 - 3}{2}\right\}, \{1, 2, 3, 2 \cdot 3, 3^2, 2 \cdot 3^2, 3^4\}\text{.}\]This still works since $3^4 = 3^3 + 2 \cdot 3^3$. Now we swap the $3^3$ on the left-hand side with the $3^2$ and $2 \cdot 3^2$ on the right-hand side again. This argument can continue until we reach the sets \[\left\{3, 2 \cdot 3, 2 \cdot 3^2, 2 \cdot 3^3, \frac{3^4 - 3}{2}\right\}, \{1, 2, 3^2, 3^3, 3^4\}\]at which point we're done. (We can actually apply the argument again, but we don't need to.)

For $n \ge 3$, consider the set $S = \{F_1, F_2, \ldots, F_{2n - 3}, x, y, z\}$, where $F_i$ is the $i$th Fibonacci number and $x$, $y$, and $z$ are integers such that $x + y + F_1 + F_2 + \cdots + F_{2n - 4} = F_{2n - 3} + z$ and none of $x$, $y$, and $z$ is equal to $F_i$ for $1 \le i \le 2n - 3$ (just make $x$ and $y$ super big). Then, if we let $\sigma$ be the sum of all elements in $S$, observe that $$\frac{\sigma}{2} = z + F_{2n - 3} = z + F_{2n - 4} + F_{2n - 5} = z + F_{2n - 4} + F_{2n - 6} + F_{2n - 7} = \cdots = \underbrace{z + \sum_{i = 1}^{n - 2}F_{2i} + F_1}_{n\text{ elements}},$$which finishes.

Consider the set $\{F_1, F_2, F_3, \dots, F_{2n-1}, \sigma - F_{2n-1}\}$ where we define $\sigma = F_1 + F_2 + \dots + F_{2n-2}$. Then consider, \begin{align*} S_2 &= \{\sigma - F_{2n - 1}, F_{2n - 1}\}\\ S_3 &= \{\sigma - F_{2n - 1}, F_{2n - 2}, F_{2n - 3}\}\\ S_4 &= \{\sigma - F_{2n - 1}, F_{2n - 2}, F_{2n - 4}, F_{2n - 5}\}\\ &\vdots\\ S_n &= \{\sigma - F_{2n - 1}, F_{2n - 2}, F_{2n - 4}, F_{2n - 6}, \dots, F_{4}, F_3 \} \end{align*}Clearly each of these partitions work so we're done. $\square$

The above solutions are too easy to read? Well no worries, I'm here For $n=3$ take the set $\left\{1,2,3,5,7,8\right\}$ with the partitions $\left\{1,2,3,7\right\}$ and $\left\{5,8\right\}$ for $m = 2$ and $\left\{2,3,8\right\}$ and $\left\{1,5,7\right\}$ for $m=3$. Now we work for $n\ge 4$. Consider the sets $S_i = \left\{2^{2n} - \sum_{k=i}^{n-1} 2^{2k}, 2^{2i} \right\}$ where $1 \le i \le n-1$. We treat these sets as arrays due to my laziness and so, the notation $S[0]$ and $S[1]$ imply the first and the second element of the set $S$. First note that $S_i[0] + S_i[1] = S_{i+1}[0]$. Now note that $\nu_{2}(S_i) = 2i$. This implies that all the sets are disjoint. Now we show that the individual elements of each set are distinct. For this note that, \[ S_i[0] = S_i[1] \iff 2^{2n} = 2^{2i} + \sum_{k=i}^{n-1} 2^{2k} \iff \nu_{2}(2^{2n}) = \nu_{2}\left(2^{2i+1} + \sum_{k=i+1}^{n-1} 2^{2k}\right) = \nu_{2}(2^{2i+1}) \iff 2n = 2i+1 \]which is clearly false. So all the all the elements of set $S_i$ are distinct. This gives us a total of $2(n-1)$ numbers. Now consider the quantity $\lambda = \sum_{i=1}^{n-1} S_i[0] + S_i[1]$. Now, \[ \nu_2(\lambda) = \nu_{2}\left(\sum_{i=1}^{n-1} S_i[0] + S_i[1]\right)= \nu_{2}\left(\left(\sum_{i=2}^{n-1} S_i[0]\right) + 2^{2n}\right) = \nu_2{2^{2\cdot 2}} = 4. \] Finally, we consider the final set $S_n = \left\{2^{2n}, \lambda - 2^{2n} \right\}$. Now using another $\nu_2$ argument (the condition $n\ge 4$ is critically important here, I'm just tired to writeup the leftover details ), it is easy to note that $\lambda -2^{2n} \not= S_2[0]$ or $S_2[1]$. This means that each of our elements of $S_i$ for $1 \le i \le n$ are all unique. Now to finish, we show that this construction works, that is $\displaystyle\bigcup_{i=1}^n S_i$ is our desired set. Firstly, note that for $m=2$, we can consider the sets $\displaystyle\bigcup_{i = 1}^{n-1} S_i$ and $S_n$. It is easy to note that both their sums are $ = \lambda$. Now for higher values of $m$, we can just swap $S_{n}[0]$ with $\left\{S_{n}[0], S_{n}[1]\right\}$ as they both have the same sum. This increases $m$ to $3$. We can similarly keep doing this by swapping $S_{i+1}[0]$ with $\left\{S_i[0], S_i[1]\right\}$ till we reach $m=n$. So done. Remark: Bro what on Earth is this solution????? It literally took me $2$ hours to rigorize the idea and then type it. . I admit that the solution looks very wordy, but the idea is pretty simple. Another thing to note that the condition $n\ge 4$ is pretty important in my solution since I keep jumping up and down the indices. For an example of the case when $n=4$, our individual sets are $S_1 = \left\{2^8 - 2^2 - 2^4 - 2^6, 2^2\right\}$, $S_2 = \left\{2^8 - 2^4 - 2^6, 2^4\right\}$, $S_3 = \left\{2^8 - 2^6, 2^6\right\}$ and $S_4 = \left\{2^8, (2^8 - 2^4 - 2^6) + (2^8 - 2^6) - 2^8\right\}$. Then our working divisions are, $\left\{(2^8 - 2^2 - 2^4 - 2^6), 2^2, (2^8 - 2^4 - 2^6), 2^4, (2^8 - 2^6), 2^6\right\} \text{ and } \left\{2^8, (2^8 - 2^4 - 2^6) + (2^8 - 2^6) - 2^8\right\}$ $\left\{(2^8 - 2^2 - 2^4 - 2^6), 2^2, (2^8 - 2^4 - 2^6), 2^4, (2^8)\right\} \text{ and } \left\{(2^8 - 2^6), 2^6, (2^8 - 2^4 - 2^6) + (2^8 - 2^6) - 2^8\right\}$ $\left\{(2^8 - 2^2 - 2^4 - 2^6), 2^2, (2^8 - 2^6), (2^8)\right\} \text{ and } \left\{(2^8 - 2^4 - 2^6), 2^4, 2^6, (2^8 - 2^4 - 2^6) + (2^8 - 2^6) - 2^8\right\}$ $\left\{(2^8 - 2^4 - 2^6), (2^8 - 2^6), (2^8)\right\} \text{ and } \left\{(2^8 - 2^2 - 2^4 - 2^6), 2^2, 2^4, 2^6, (2^8 - 2^4 - 2^6) + (2^8 - 2^6) - 2^8\right\}.$

I'm really confused.... does this work? Take $a_i=N+i$ for $0 \leq i \leq n-2$ and some $i$. Let $s_i=\sum_{k=0}^{i}a_i$ for $1 \leq i \leq n-2$. I claim the set \[\{a_0,a_1,\dots, a_{n-2}, s_1, s_2, \dots, s_{n-2}, X_1, X_2, C\}\]for sufficiently large $N$ and $C$ such that $C+s_{n-2}$ is half the sum of the whole set works. Indeed, the subsets that are half the sum of the whole thing are \[ \{C, s_{n-2}\}, \{C, s_{n-3}, a_{n-2}\}, \{C, s_{n-4}, a_{n-2}, a_{n-3}\}, \dots, \{C, s_{1}, a_{n-2}, a_{n-3}, \dots, a_2\}, \{C, a_{n-2}, a_{n-1}, \dots, a_{0}\} \]

We prove this over rationals, since we can just multiply common denominators to get an integer solution. We use induction, we show that if we can find a set for $2n$ elements we can find one for $2n + 4$. Base Case: $1,3,5,7$ for $n = 2$, and $1,1,2,2,2,4$ for $n = 3$. Inductive Step: Let the sum of all the elements be $s$. We can let the first $2n$ elements sum to $\frac s8$ and be a scaled version of an already working set with $2n$ elements, then let the other $4$ elements be $\frac s8, \frac s4, \frac{s}{16}, \frac{7s}{16}$. Then for $m = 2$, we can take $\frac{s}{16}, \frac{7s}{16}$ with sum $\frac s2$. For $2 < m < n + 2$, we can take the set with $m - 1$ elements with sum $\frac{s}{16}$ from the first $2n$ elements and then add $\frac{7s}{16}$. Lastly, for $m = n + 2$, we can take the set with $n - 1$ elements with sum $\frac{s}{16}$ from the first $2n$ elements, then add $\frac{s}{16}, \frac s8, \frac s4$, thus we can find all of the desired subsets. It just remains to show that this construction is actually a set. We show that none of the first $2n$ elements have value at least $\frac{s}{16}$, this is obvious since we could not possibly have a number that large, since no set containing it could sum to half of $\frac s8$ and no set excluding it and at least one other element could sum to $\frac s8$. Thus every element of the first $2n$ elements are less than the $4$ elements at the end, none of the $4$ elements at the end are equal, and none of the first $2n$ elements are equal by the inductive hypothesis, so we are done.

THIS TOOK ME 80 MINUTES We use the Fibonacci sequence with the indexing $F_0=F_1=1$. Specifically, we set $S=\{F_1,F_2,\dots,F_{2n-1}\}\cup\{F_{2n-2}-2\}$. For any $m\in\{2,3,\dots,n\}$, one valid partition is then given by $S=X\cup Y$, where \[U=\{F_{2n-2m+4},F_{2n-2m+6},\dots,F_{2n-2}\}\cup\{F_{2n-2m+3},F_{2n-2}-2\}\]and $Y=S\setminus X$. It is easy to verify that this partition works by inducting on $m$ and also observing the identity $\sum_{i=1}^kF_i=F_{k+2}-2$.

Consider the set $\{1, 2, 2^2, 2^3,... 2^{n-1}, 2^n-1, 2^{n-1}-1, ..., 2^2-1, k\}$, where $k$ is any integer different from the ones already in the set. It is easy to notice that the sum of the first $m$(where $2 \le m \le n$) elements from the left is actually the $m^{th}$ element from right.