Let $m$ and $n$ be positive integers and $p$ be a prime number. Find the greatest positive integer $s$ (as a function of $m,n$ and $p$) such that from a random set of $mnp$ positive integers we can choose $snp$ numbers, such that they can be partitioned into $s$ sets of $np$ numbers, such that the sum of the numbers in every group gives the same remainder when divided by $p$.
1. **Understanding the Problem:**
   We need to find the greatest positive integer \( s \) such that from a random set of \( mnp \) positive integers, we can choose \( snp \) numbers, which can be partitioned into \( s \) sets of \( np \) numbers, where the sum of the numbers in each set gives the same remainder when divided by \( p \).

2. **Applying the Erdős–Ginzburg–Ziv Theorem:**
   The Erdős–Ginzburg–Ziv theorem states that for any set of \( 2p-1 \) integers, there exists a subset of \( p \) integers whose sum is congruent to \( 0 \pmod{p} \). This theorem is crucial for our solution.

3. **Using the Theorem:**
   Given \( mnp \) integers, we need to partition them into \( s \) sets of \( np \) integers each. According to the theorem, for any set of \( 2p-1 \) integers, we can find \( p \) integers whose sum is \( 0 \pmod{p} \). 

4. **Partitioning the Integers:**
   To ensure that we can partition \( snp \) integers into \( s \) sets of \( np \) integers each, where each set's sum gives the same remainder modulo \( p \), we need to consider the constraints given by the Erdős–Ginzburg–Ziv theorem.

5. **Calculating \( s \):**
   Since we need \( snp \) integers to be partitioned into \( s \) sets of \( np \) integers, and each set of \( np \) integers must satisfy the condition given by the theorem, we need to ensure that \( snp \leq mnp \). This simplifies to \( s \leq m \).

6. **Conclusion:**
   The greatest positive integer \( s \) such that we can partition \( snp \) integers into \( s \) sets of \( np \) integers, each set having the same remainder when divided by \( p \), is \( s = m \).

The final answer is \( \boxed{m} \).