Problem: Determine all positive integers $n\geq 2$ for which there exist integers
$x_1$, $x_2$, \dots, $x_{n-1}$ satisfying the condition that
if $0<i<n$, $0<j<n$, $i\neq j$ and $n$ divides $2i+j$, then $x_i<x_j$.

Solution: 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:
\begin{itemize}
  \ii 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
  \[ 3 \cdot 2^{i-1} \cdot t_1 \equiv 0 \pmod n \qquad (\heartsuit). \]
  \ii For $t_m \prec t_1$ to be true, we need
  \[ (-2)^m t_1 \equiv t_1 \pmod n \iff \left( (-2)^m-1 \right) t_1 \equiv 0 \pmod n.
    \qquad (\spadesuit) \]
\end{itemize}
We now analyze three cases:
\begin{itemize}
  \ii 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.

  \ii 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.

  \ii 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.
\end{itemize}
Thus these are all the answers and the only answers.