We use the notation $x_{n+i}=x_i$. Let $a_k$ be the sequence $a_k = \sum_{i=1}^n\left|x_{i+1,k}-x_{i,k}\right|$. This sequence is non-negative and monotonically decreasing, as $a_{k+1}= \sum_{i=1}^n\left|\frac{x_{i+2,k}+x_{i+1,k}}{2}-\frac{x_{i+1,k}+x_{i,k}}{2}\right|$ $\leq\frac12\sum_{i=1}^n\left(|x_{i+2,k}-x_{i+1,k}|+|x_{i+1,k}-x_{i,k}|\right) = \sum_{i=1}^n|x_{i+1,k}-x_{i,k}| = a_k$. Equality holds iff $x_{1,k}=x_{2,k}=\cdots=x_{n,k}$. In case of odd $n$, $x_{1,k+1}=x_{2,k+1}=\cdots=x_{n,k+1}$ implies $x_{1,k}=x_{3,k}=\cdots=x_{n,k} = x_{2,k} = \cdots = x_{n-1,k}$, so $a_{k+1}=0 \implies a_{k}=0$. (That's not the case if $n$ is even, then $x_{2i,1}=l,x_{2i+1,1}=2m-l$ is a counterexample). Not all $x_{i,1}$ are equal, hence $a_1>0$ and the sequence $a_k$ is strictly positive and strictly monotonically decreasing, therefore $a_k\not\in\mathbb{Z}$ for large values of $k$, which implies that for a large enough $k$ there is an index $i$ such that $a_{k,i}\not\in\mathbb{Z}$.