Croatian Mathematical Competitions

Let us assume that $k$ does not divide $n$ and that $n = kq + r$, $0 < r < k$.
Since only the remainder of division of $m$ by $k$ is relevant, without loss of generality we may assume $0 < m \le k$. We will prove that $m = k$.
If we assume that $m \le r$, then the numbers $p_m, p_{m+k}, \dots, p_{m+kw}$ would be equal to the numbers $k, 2k, \dots, qk$ in some order. This is impossible since the first sequence has $q+1$ elements, and the second sequence has $q$ elements.
Hence $m > r$ and $m + 1 + qk > n$. There are $q+1$ numbers $1, k+1, \dots, qk+1$ which are equal to numbers $p_{m+1-k}, p_{m+1}, p_{m+1+k}, \dots, p_{m+1+(q-1)k}$. This implies $m+1-k \ge 1$, i.e. $m \ge k$. Since we assumed that $m \le k$ it follows that $m = k$.