Suppose, for the sake of contradiction, that all of the $x_{j, k}$ were integers. First, we claim there does not exist a $k$ with $x_{1, k} = x_{2, k} = x_{3, k} = \dots = x_{n, k}$. If there did, consider the smallest one (note we are given $k \ne 1$). By definition, \[x_{1, k} = \frac{x_{1, k - 1} + x_{2, k - 1}}{2}\] \[x_{2, k} = \frac{x_{2, k - 1} + x_{3, k - 1}}{2}\] and so on, which forces $x_{1, k - 1} = x_{3, k - 1} = x_{5, k - 1} = \dots = x_{n, k - 1} = x_{2, k - 1} = x_{4, k - 1} = \dots x_{n -1, k - 1}$, contradicting the minimality of $k$. Let $m$ be the smallest integer such that $x_{j, 1} \le m$ for all $j$. Since not all the $x_{j, 1}$ are equal, there exists some $k$ with $x_{j, k} \le m - 1$. Then \[x_{j - 1, k + 1} = \frac{x_{j - 1, k} + x_{j, k}}{2} \le \frac{2m - 1}{2}\] \[x_{j, k + 1} = \frac{x_{j, k} + x_{j + 1, k}}{2} \le \frac{2m - 1}{2}\] but $x_{j - 1, k + 1}, x_{j, k + 1}$ are integers by hypothesis, so they are each at most $m - 1$. Repeating this, for all $k$ large enough the $x_{j, k}$ are at most $m - 1$. We can then repeat the argument, and for $k$ large enough all the $x_{j, k}$ are at most $m - 2$. Then repeating as necessary, for $k$ sufficiently large, all the $x_{j, k}$ ($1 \le j \le n$) are all arbitrarily small, contradicting the fact that their sum is $x_{1, 1} + x_{2, 1} + \dots + x_{n, 1}$ which is fixed. Thus there exist indices $j, k$ for which $x_{j, k}$ is not an integer. When $n$ is even, this isn't true; we could take our initial sequence of integers to be $1, -1, 1, -1, \dots, 1, -1$ whence all the $x_{j, k}$ are integers.