Let $n=2^s$ And let $X_{i}=v_{2}(i)$. Then clearly if $n|2i+j$ if $v_{2}(j)\le v_{2}(i)$ then $v_{2}(j)=s$ but $0<j<2^s$ a contradiction. So $n=2^s$ satisfies. Let $n=2^s*b$ where $b>3$ is odd. Now consider $X_{2^s},X_{2^s*2},X_{2^s*3},...,X_{2^s*(b-1)}$. Then since $n|2i+j$ where $i,j$ are from $2^s,2^s*2,2^s*3,...,2^s*(b-1)$ is equivalent to $b|2x+y$ where $i=2^s*x,j=2^s*y$ But for each $x$ from $0,1,2,3,..,b-1$ there is a unique $y=p(x)$ from $0,1,2...,b-1$ such that $b|2x+y$ since $(2,b)=1$. Now $p(0)=0$ so $p(1),p(2),...,p(b-1)$ is a permutation of $1,2,3...,b-1$ therefore getting back to the original problem using the condition for $X_{2^s},X_{2^s*2},X_{2^s*3},...,X_{2^s*(b-1)}$ we get that $X_{2^s}+X_{2^s*2}+X_{2^s*3}+...+X_{2^s*(b-1)}>X_{2^s}+X_{2^s*2}+X_{2^s*3}+...+X_{2^s*(b-1)}$ which is a contradiction. Answer is all powers of $2$.

leader, you seem to have forgotten about the case $n = 3 \cdot 2^k$. A more intuitive way to think about this is that we simply do not want the existence of cycles. First look at $n$ odd and $\ge 5$. For a cycle to exists, say $a_1, a_2, \ldots, a_j$ (these are indices from $1$ to $n-1$) are in the cycle. Then $a_{k+1} = -2{a_k}$ for all $k$, so we can just set $j$ to be the order of $-2 \pmod{n}$. Note that $n > 3$ is important or else $a_k = a_{k+1}$. For $n = o \cdot 2^k$, $o \ge 5$ just look only at the indices that are multiples of $2^k$ and repeat the construction for $o$ on them. Proving that there are no cycles for $n = 2^k$ is done above. For $n = 3 \cdot 2^k$, it is a similar argument as above (what leader did to $n = 2^k$) but more caseworky. (You have to directly check that the indices $2^k, 2 \cdot 2^k$ never appear, but this is relatively easy)

Ah yes I made a mistake because i did not include that condition $i\not= j$ since for that case we have $X_{i]=X_{j}}$ so odd numbers are possible whenever $n|2i+j$ implies $i=j$ for so this means $n|3$ so $3$ is also possible.

Dear mathlinkers, From my point of view some of $x_1,x_2,...,x_{n-1}$ can be equal...

In case $n$ is odd, then $n=2k+1$ and $(i;j)=(1;2k-1),(2;2k-3),(3;2k-5),...,(k;1),(k+1;2k),(k+2;2k-2),..., (2k-1;4),(2k;2)$ then $x_1<x_{2k-1},...,x_k<x_1, x_{k+1}<x_{2k},...,x_{2k}<x_2$ Then we have their sum respectively : $\sum^{2k}_{i=1}x_i<\sum^{2k}_{i=1}x_i$, contradiction! But $n=3$ we have $i=3$ thus it can be solution => $n$ is even If $n=4k+2$ then that is sufficient to have contradiction $(i;j)=(2;4k-2),(4;4k-6),...,(2k;2),(2k+2;4k),...,(4k;4)$ then we have $\sum^{2k}_{i=1}x_{2i}<\sum^{2k}_{i=1}x_{2i}$ . But in case $n=6$ we have $i=j$ in above properties thus $n=6$ can also be solution If $n=4k$ then we put $x_i=1$ if $i$ is odd , $x_i=2$ if $n=4k+2$ and $x_i=4$ if $n=4k$, this can be solution. Answer is $n=3;6;4k$ for all $k=1,2,...$!

Take n 20, i 16 and j 8. Then x_i=x_j , but this cant according to the conditions, hence isho, you have a mistake in your third case. And obviously n = 2 worked too btw.

Ah, sorry I understood my mistake.

this is the official solution

Attachments:

Solution from Twitch Solves ISL: The answer is $n = 2^k$ and $n = 3 \cdot 2^k$, for each $k \ge 0$ (excluding $n=1$). We work with the set $S = \{1, 2, \dots, n-1\} \bmod n$ of nonzero residues modulo $n$ instead. We define the relation $\prec$ on $S$ to mean that $2i + j \equiv 0 \pmod n$ and $i \neq j$, for $i,j \in S$. Then the problem satisfies the conditions if and only if $\prec$ has no cycles, i.e.\ $\prec$ imposes a partial order on $S$. The existence of a cycle for $\prec$ is equivalent to some choice of $t_1 \in S$ and an integer $m \ge 2$ such that \[ t_1 \prec t_2 \prec \dots \prec t_m \prec t_1. \]Unwinding the definition, this is equivalent to two conditions: We need $t_i \not\equiv t_{i+1} \pmod n$ for $i=1,\dots,m$ (where $t_{m+1}=t_1$). This is equivalent to \[ 3 \cdot 2^{i-1} \cdot t_1 \equiv 0 \pmod n \qquad (\heartsuit). \] For $t_m \prec t_1$ to be true, we need \[ (-2)^m t_1 \equiv t_1 \pmod n \iff \left( (-2)^m-1 \right) t_1 \equiv 0 \pmod n. \qquad (\spadesuit) \] We now analyze three cases: Let $n = 2^k$. Suppose for contradiction some cycle exists. Then $(-2)^m-1$ is coprime to $n$, so $(\spadesuit)$ would imply $t_1 \equiv 0 \pmod n$, contradiction. Let $n = 3 \cdot 2^k$. Suppose for contradiction some cycle exists. If $(\spadesuit)$ holds for some $m$, then $2^k \mid t_1$, so the only possibility is that $t_1 \equiv \pm 2^k \pmod{n}$ and $3 \mid (-2)^m-1$. However, in that case $(\heartsuit)$ is violated for $i=1$, contradiction. Suppose $n$ had an $n$ has an odd divisor $d \mid n$ and $d \ge 5$. Then taking $t_1 = n/d$ and $m = \varphi(d)$, the equation $(\spadesuit)$ is true. Moreover, $(\heartsuit)$ is true because there is at least one odd prime $p$ with $\nu_p(n) > \nu_p(3t_1) = \nu_p(3n/d)$ (since $d \ge 5$ is odd). So indeed it's possible to construct a cycle. Thus these are all the answers and the only answers.