, 2022

$s = m - 1$. Assume $s = m$ and consider a set of $mnp - 1$ positive integers, congruent $1 \pmod{p}$, and $p$. Clearly this set does not fulfill the statement, thus $s \le m - 1$.

Lemma. Among every $np + p - 1$ positive integers, there exist $np$ with sum, divisible by $p$.

Proof. We apply induction to $n$. For the base $n = 1$ we have to show, that among every $2p - 1$ positive integers, there exist $p$ with sum, divisible by $p$. Consider residues modulo $p$, and via induction on $k$, prove that for all $p \ge k \ge 2$ the set of the residues $S$ of all sums of $k$ elements of an arbitrary $(2k-1)$-element set that does not contain $k$ equal elements, satisfies $|S| \ge k$. For $k = 2$ the statement is obvious. Assume, we proved it for $k \le p-1$ and let $s_1, s_2, \dots, s_k$ be the different residues. Consider a $2k+1$-element set and two different elements in it $a$ and $b$. Apply the inductive hypothesis to the remaining subset (it is clear that we can choose $a$ and $b$ such that, the remaining subset be admissible). Consider the sets $\{s_1 + a, s_2 + a, \dots, s_k + a\}$ and $\{s_1 + b, s_2 + b, \dots, s_k + b\}$. If they do not coincide, we derive $k+1$ different residues and the auxiliary induction is completed. Assuming the sets coincide, we derive after summation that $a \equiv b \pmod{p}$ - a contradiction. Hence, we have proven the base $n = 1$ for our major induction. Finally, if we have constructed our $(n-1)p$-element subset with sum, divisible by $p$, then among the remaining $2p-1$ numbers there are $p$ with sum, divisible by $p$, and we can add them to the subset. The lemma is proved.

Now, applying consecutively the lemma, and extracting new $np$ elements with sum $0 \pmod{p}$ at each step, we will end up with $m-1$ disjoint subsets with sum of the elements $0 \pmod{p}$, so the proof is completed.