Note that I read it as: You can find a sequence so that no arithmetic sequence is an integer. My solution still is correct for the problem though.

yugrey wrote: For $n \ge 4$, the first $n-2$ terms and similarly the last $n-2$ have sum at most $2n-5$ Why? I see asolutely no reason, why would sum of first $k$ numbers in any way limit the sum of the last. Or am I missing something? And by the way, how did you come up with this idea? It's so unintuitive...

Note the "similarly." It is the exact same argument.

Note that if the sum for some $k$ consecutive numbers of the sequence is $2k$ we are finished. Now it can't be $a_1+a_2+\dots+a_{n-1}=2n-2$ so we conclude $a_1+a_2+\dots+a_{n-1}\le 2n-3$. Now we have that $a_1+\dots + a_{n-1}>a_1+\dots+a_{n-2}$ so we conclude $a_1+a_2+\dots +a_{n-2} \le 2n-5$. By continuing like this we get $a_1+a_2 \le 3$. Now for $n=2$ and $n=3$ examples are $ 2 1 $ and $2 1 2$. For $n=4$, $n=5$ we can easily show that there has to be an arithmetic mean that is an integer. Now for $n\ge6$ if $a_1+a_2=2$ we are finished and if $a_1+a_2=3$ then $a_3+\dots+a_n=2n-4$ we are finished by induction(because $n-2\ge4$). So the answer is for all $n\ge4$. My solution is very similar to yugrey's and I think it is very intuitive because you can just notice these things. He just wrote it more professional.

Unless I made some mistake it's neither casebashy nor hard.

I think this is different from aboves:

Groupsolve with TOMITTAL and Mathotsav First we replace $a_i$ by $a_i-1$ to get a new equivalent problem- New Probelem wrote: Determine all positive integers $n$, $n\ge2$, such that the following statement is true: If $(a_1,a_2,...,a_n)$ is a sequence of non-negative integers with $a_1+a_2+\cdots+a_n=n-1$, then there is block of (at least two) consecutive terms in the sequence with their (arithmetic) mean being an integer. Solution- We will show that it is true for all $n>3$ For $n=2,3$ the following sequences are counter examples - $(1,0),(1,0,1)$ Now we will prove a more general result - Result- wrote: For $n>1$ if we have a sequence $(a_1,a_2,....,a_n)$ of non-negative integers with $a_1+a_2+....+a_n=n-k$ where $0\leq k \leq \lfloor \dfrac{n}2 \rfloor-1 $. Then there exists a block of $l$ consecutive terms with sum $l$ for some $1<l\leq n$ Proof- Assume for the sake of Contradiction that such a block does not exist. Define $b_0=0$ and $b_i=i-(a_1+a_2+...+a_i)$ Claim 1 - $b_{i+1}-b_i\leq 1$ $b_{i+1}-b_i=1-a_{i+1}\leq 1$ Claim 2 - If $b_i=x$ and $b_j=y$ and $i<j , x<y$ then all values between $x$ and $y$ are taken by some $b_t$ for $i<t<j$. Follows directly from Claim 1 Claim 3 - If $b_i=b_j$ then $|i-j|\leq 1$ Let $b_i=b_j$ and $j>i$ , so $b_j-b_i=0 \implies (j-i) - (a_{i+1}+...+a_j) = 0 \implies (a_{i+1}+...+a_j)=j-i $ And so by our assumption that such a block of lenght $>1$ cannot exist we get that $j-i=1$ Claim 4 - $b_i$ is non-decreasing. We know that $b_n=k \geq 0 =b_0$ and hence if $b_i$ is decreasing at some point then it must be increasing at some other point. So we get $i<j<k$ such that $a_i,a_k>a_j$ , now by claim 2 there exists $s,t$ , such that $i\leq s<j<t\leq k$ and $a_s=a_j-1=a_t$ , contradicting claim 3. Thus we get that the $b_i$ are all from $\{0,1,..,k\}$ , but from claim 3 we have that each number can only occur twice. And hence $n+1 \leq 2(k+1) \implies n<2k+2 \implies \lfloor \dfrac{n}2 \rfloor<k+1$ Contradiction, Hence proved. Note : The bound on $k$ is strong . This can be seen by considering the sequence $(1,0,1,0,...)$ Further see that $1\leq \lfloor \dfrac{n}2 \rfloor-1$ holds only for $n>3$ and hence our original problem is also solved.

The answer is all $n$ except $n = 2$ and $n = 3$, for which we may pick the sequences $2, 1$ and $2, 1, 2$. We say a sequence $(a_1, \ldots , a_n)$ is exceptional if there is no block of consecutive terms whose average is an integer. We prove that for $n \ge 4$ there is no exceptional sequence whose elements have sum $2n-1$ by proving the following stronger result. Claim. Let $(a_1, \ldots , a_n)$ be an exceptional sequence. Assume $a_1 + \ldots + a_n \le 2n-1$. Then $a_1 + \ldots + a_n \le \frac{3n+1}{2}$ (i.e. the sequence is of the form $1, 2, 1, 2, \ldots$ or $2, 1, 2, 1, \ldots$). We prove the claim by strong induction on $n$. The base cases $n = 1, 2, 3$ are trivial. Assume $n \ge 4$. For the induction step, first note that if $a_1 + \ldots + a_n \le 2n-1$, then for any $k$ we have either $a_1 + \ldots + a_k \le 2k-1$ or $a_{k+1} + \ldots + a_n \le 2(n - k) - 1$. Apply this for $k = \lfloor n/2 \rfloor$ and abuse symmetry to deduce that $a_1 + \ldots + a_{\lfloor n/2 \rfloor} \le 2\lfloor n/2 \rfloor - 1$. By the induction assumption this implies that the numbers $a_1, a_2, \ldots , a_{\lfloor n/2 \rfloor}$ alternate between $1$ and $2$. We now assume that $a_1, a_2, \ldots$ does not alternate between $1$ and $2$ and aim for contradiction. Let $m$ be the smallest integer such that $a_m \not\in \{1, 2\}$, so $m \ge \lfloor n/2 \rfloor + 1 \ge 3$. The crucial observation is that $a_m$ must be quite large. We first prove a weaker result in this direction and the prove a stronger result which is sufficient for a contradiction. The first result is. \begin{align*} a_m \ge \lfloor \frac{m}{2} \rfloor + 3. \end{align*}Let $k \ge 1$ be an integer such that $2k \le m$. Now \begin{align*} \frac{a_{m - 2k} + a_{m - 2k + 1} + \ldots + a_{m-1} + a_m}{2k+1} = \frac{3k + a_m}{2k+1} = 1 + \frac{k-1 + a_m}{2k+1}, \end{align*}so $a_m \neq k+2$. Applying this for $k = 1, 2, \ldots , \lfloor \frac{m}{2} \rfloor$ yields the desired lower bound for $a_m$. We then note that since $a_1 + \ldots + a_{m-1} \ge \frac{3(m-1) - 1}{2}$, we have \begin{align*} a_{m+1} + \ldots + a_n = (a_1 + \ldots + a_n) - (a_1 + \ldots + a_m) \le 2n - 1 - \left(\frac{3(m-1) - 1}{2} + \lfloor \frac{m}{2} \rfloor + 3\right) \le 2(n-m) - 1, \end{align*}so by the induction hypothesis the numbers $a_{m+1}, \ldots , a_n$ must alternate between $1$ and $2$. Furthermore, $a_{m+1} = a_{m-1}$ by considering parities. We now prove a stronger lower bound for $a_m$ than the one before, namely that \begin{align*} a_m \ge \lfloor \frac{n}{2} \rfloor + 1. \end{align*}We use a similar trick as for the previous lower bound but "in two directions". Let $e(m) = 0$ if $a_{m-1} = 1$ and $e(m) = 1$ if $a_{m-1} = 2$. For $k \ge 1$ an integer with $2k \le n$ we have \begin{align*} \frac{a_1 + a_2 + \ldots + a_{2k}}{2k} = \frac{3k - e(m) + a_m}{2k} = 1 + \frac{k - e(m) + a_m}{2k}, \end{align*}and hence $a_m \neq k + e(m)$. Applying this for $k = 1, 2, \ldots , \lfloor n/2 \rfloor$ gives the stronger lower bound for $a_m$. Finally, use the obtained bound to get \begin{align*} a_1 + \ldots + a_n = (a_1 + \ldots + a_{m-1}) + a_m + (a_{m+1} + \ldots + a_n) \ge \frac{3(m-1) - 1}{2} + \lfloor \frac{n}{2} \rfloor + 1 + \frac{3(n - m) - 1}{2} = \frac{3(n-1)}{2} + \lfloor \frac{n}{2} \rfloor > \frac{3(n-1)}{2} + \frac{n}{2} = 2n - 3/2. \end{align*}Hence $a_1 + \ldots + a_n \ge 2n-1$. This is almost enough. To get rid of the equality case, note that $a_m = \lfloor n/2 \rfloor + 1$ can happen only if $e(m) = 1$, i.e. if $a_{m-1} = 2$. In this case $a_1 + \ldots + a_{m-1} \ge 3(m-1)/2$ and $a_{m+1} + \ldots + a_n \ge 3(n-m)/2$, yielding an additional $+1$ term for the lower bound. Remark. A simpler proof of the claim can be given by an easy casework on the values of $a_1$, see post #8.

Solution from stream: The answer is all $n\ge 4$. Note $n=2,3$ fail because of $(1,2)$, $(2,1,2)$ respectively. Assume for contradiction for some $n\ge 4$, there exist a sequence that doesn't satisfy the statement in the question. Lemma 1. For all integers $k\ge 1, \sum\limits_{j=i+1}^{i+2^k} a_j \equiv 2^{k-1} (\bmod\; 2^k)$ Proof. We proceed by induction on $k$. The base case, $k=1$ is true because otherwise $2\mid a_i+a_{i+1}$ results in a contradiction. Inductive Step: We consider $\sum\limits_{j=i+1}^{i+2^{k-1}} a_j \equiv \{ 2^{k-2}, -2^{k-2} \} (\bmod\; 2^k)$, which is a restatement of the inductive hypothesis on $k-1$. Now, note $$\sum\limits_{j=i+1}^{i+2^{k-1}} a_j + \sum\limits_{j=i+1+2^{k-1}}^{i+2^k} a_j = \sum\limits_{j=i+1}^{i+2^k} a_j$$cannot be divisible by $2^k$, so we are good. A corollary of this lemma is that $a_i\equiv a_{i+2^k} (\bmod\; 2^k)$. Consider a sequence $b_i$ with alternating $2$'s and $1$'s with odd sum (if $4\mid n$, then we clearly lose because we can group $a_i$ into $\frac n2$ groups with odd sum, which has an even total sum). Notice $$\sum b_i=\frac{3n - Im(i^n)}{2}\ge \frac{3n-1}{2}$$ Now, suppose $a_i>b_i$ for some $i$. Obviously, $a_i\equiv b_i(\bmod\; 2)$ (in the case where $n$ is 2 mod 4, we can reverse the sequence to make this true) Let $k=\nu_2(a_i-b_i)$. Observe that the sum of any continuous $2^{k+1}$ elements must have incremented by at least $2^{k+1}$ for mod $2^{k+1}$ reasons by our lemma. We can easily bound to prove $\sum\limits_{i=1}^n (a_i-b_i) \ge 2^{k+1} \max\{\lfloor \frac{n}{2^{k+1}}\rfloor, 1\}$. If $2^{k+1}>\frac n2$ we're trivially done, otherwise, $2^{k+1} \lfloor \frac{n}{2^{k+1}}\rfloor \ge n-2^{k+1} > \frac{n-1}{2}$ for all $n\ge 5$, so we win. Motivational Remarks: Note alternating sequences of two consecutive numbers work if the sum condition is not given. This is an algebraic estimation/bounding problem, because the $\sum a_i=2n-1$ is also a strong condition. Note the $(1,2,1,2,)$'s serve as a basis, for $a_i$ is at least $b_i$ for all $i$. Now, note $2\nmid a_i+a_{i+1}$ is a very STRONG condition, which gives away a lot about parity information. Then we can extend to $4\nmid a_i+a_{i+1}+a_{i+2}+a_{i+3}$ and so on, so on. The equality cases may also be a good starting place.

n≠1. Counter examples: n=2 - (1,2) n=3 - (2,1,2) Let n≥4, n can be 4k, 4k+1, 4k+2, 4k+3 for some k≥1. Let us construct the sequence where no blocks exists with the required property and call them 'rebel'. For any two terms a,b, (a+b)/2 to be an integer a, b must have opposite parity. (a,b) can be (odd, even) or (even,odd) Consider the first 4 terms: (odd, even, odd, even) or (even, odd, even, odd) Let (a,b,c,d) is denoted by abcd. So first 4 terms mod 4 are: 1010 1012 1212 10-10 10-12 12-12 -10-10 -10-12 -12-12 We can't have the sum 0 mod 4, as then their means will be an integer. We are left with the possiblities: 1010 10-12 -10-10 Observe that the full sequence is the concatenation of these terms + (1 or 2 or 3) terms. Consider the case where all evens are 0 mod 4. => their minimum value = 4. So the minimum of the sequence = 4×(no. of. min. even terms) + sigma odds = 4(n-1)/2 + sigma odds = 2n-2 + sigma odds > sigma sequence = 2n-1 (given). A Contradiction. Hence we are left with the possiblilty 10-12. The min sum of first 4 terms = 1+2+3+4 = 10. => Possible sum of the sequence = (10k or 10k+1 or 10k+3 or 10k+6) + sigma odds > Sum of the sequence = 2n-1 = (8k-1, 8k+1, 8k+3, 8k+5) Therefore, there is no 'rebel' and the question statement is true for all n≥4. (ANY MISTAKES?)

The answer is : $n \geq 4$ We can easily see that for $n=2$ and $n=3$ $(1,2)$ and $(2,1,2)$ contradicts. Assume contrary for $n\geq 4$ Claim:$a_1+a_2+...+a_i \leq 2i$ Proof:Take maximal $i$ doesn't obey the claim.Obviously $i\not =n$ We have $a_1+a_2+...a_i \geq2i+1$ and $a_1+a_2+...a_i+a_{i+1} \leq 2i+2$ $\rightarrow$ $a_1+a_2+...a_i+a_{i+1} =2i+2$ .Contradiction. Then $a_1\leq 2$ if $a_1=1$ then $n-1|a_2+a_3+...+a_n$ contradiction.If $a_1=2$ then $a_1+a_2 \leq 4$ if $a_2=2$ $2|a_1+a_2$ contadiction . If $a_2=1$ then $n-2|a_3+a_4+...+a_n$ contradiction . $\blacksquare$

The answer is all integers larger than $3$. When $n=2$, take $(a_1,a_2)=(1,2)$. When $n=3$, take $(a_1,a_2,a_3)=(2,1,2)$. Check that no block of consecutive terms have integer arithmetic means. Now we give the following claim, which is in fact a stronger result. Claim: Assume $n$ is an integer larger than $3$, and $a_1,\dots,a_n$ are $n$ positive integers with sum $2n-1$. There exist integer $1\leq i<j\leq n$, such that $\sum\limits_{k=i}^{j}a_k=2(j-i+1)$ Proof of the claim: For $1\leq k \leq n$, define $S_k=\sum\limits_{t=1}^{k}a_t$, also set $S_0=0$. Therefore $S_n=-1$, and note that $S_{k}-S_{k-1}=a_{k}-2\geq -1$ for $1\leq k\leq n$. Our goal is to find $S_i=S_j$ with $0\leq i<i+1<j\leq n$. Since $S_{n-1}\leq S_n+1=0$, we have the following cases. Case 1: $S_{n-1}=0$ $S_0=S_{n-1}=0$, done. Case 2: $S_{n-1}<-1$ Because $S_0=0$ and $S_{n-1}<-1$, there exists $1\leq i<n-1$ such that $S_i=-1$. Thus $S_i=S_n=-1$, done. Case 3: $S_{n-1}=-1$ In this case $S_{n-2}\leq S_{n-1}+1=0$, we have three subcases. Subcase 1: $S_{n-2}=0$ $S_{n-2}=S_0=0$ and $n-2\geq 2$, done. Subcase 2: $S_{n-2}=-1$ $S_{n-2}=S_n=-1$, done! Subcase 3: $S_{n-2}<-1$ Analogously there exists $1\leq i<n-2$ such that $S_i=-1$, hence $S_i=S_n$, done. We have exausted all cases, therefore the claim is proved. $\blacksquare$ Remark: The idea of this solution is somehow similar to https://artofproblemsolving.com/community/c6h407372/url]