Problem 966

Prove that $x_{i}^{(k)}=x_{i}^{(k-1)}+x_{i+1}^{(k-1)}$, hence by induction it is easy to get
$$x_{i}^{(k)}=\sum_{i=0}^{k} \mathrm{C}_{i}^{i} x_{i+1}^{(0)}(\bmod 2)$$

For simplicity, the following derivation omits $\bmod 2$, and let $x_{i+1}^{(0)}=x_{i+1}$.
If $m$ is not a multiple of $n$, then $m=q n+r, 0<r<n$, let
$$a_{i}=\left\{\begin{array}{l} \mathrm{C}_{m}^{i}+\mathrm{C}_{m}^{i+n}+\cdots+\mathrm{C}_{m}^{i+q n}, 1 \leqslant i \leqslant r \\ \mathrm{C}_{m}^{i}+\mathrm{C}_{m}^{i+n}+\cdots+\mathrm{C}_{m}^{i+(q-1)}, r<i \end{array}\right.$$

Then
$$x_{i}^{(m)}=x_{i}+a_{1} x_{i+1}+a_{2} x_{i+2}+\cdots+a_{n} x_{i+n}$$

Hence
$$a_{1} x_{i+1}+a_{2} x_{i+2}+\cdots+a_{n} x_{i+n}=0$$

Let $i=1,2, \cdots, n$, then we have
$$a_{n-1}+a_{n}=0, a_{n-2}+a_{n-1}=0, \cdots, a_{1}+a_{n}=0$$

Thus $a_{1}, a_{2}, \cdots, a_{n}$ have the same parity.
On the other hand,
$$\begin{aligned} a_{r}= & \mathrm{C}_{m}^{r}+\mathrm{C}_{m}^{r+n}+\cdots+\mathrm{C}_{m}^{r+q n}= \\ & \mathrm{C}_{m}^{m-r}+\mathrm{C}_{m}^{m-r-n}+\cdots+\mathrm{C}_{m}^{m-r-q n}= \\ & \mathrm{C}_{m}^{q n}+\mathrm{C}_{m}^{(q-1) n}+\cdots+\mathrm{C}_{m}^{n}+1= \\ & a_{n}+1 \end{aligned}$$

This shows that $a_{r}$ and $a_{n}$ have different parities, which is a contradiction. Therefore, $m$ must be a multiple of $n$.