, 2022

Without loss of generality we can consider the elements of $M$ as positive integers, as we can multiply them by the $\gcd$ of their denominators, which does not affect the problem statement.

The arithmetic progression $\{(2n-1)q+n\}_{q=1}^{\infty}$ contains infinitely many primes, due to the Dirichlet theorem and $((2n-1), n) = 1$. Let us fix a prime $p = (2n-1)q+n$, larger than all the elements of $M$, i.e., $M$ becomes a set of residues modulo $p$. Consider the following set of residues:

$$P = \left\{ \frac{p+n-1}{2n-1}, \frac{p+3n-2}{2n-1}, \dots, \frac{2p-1}{2n-1} \right\},$$

for which the numerator of every successive fraction exceeds the previous one by $2n-1$. According to the choice of $p$, all elements in $P$ are positive integers. Moreover
$$2 \cdot \frac{p+n-1}{2n-1} = \frac{2p-1}{2n-1} + 1 \quad \text{and} \quad n \cdot \frac{2p-1}{2n-1} = p + \frac{p-n}{2n-1}.$$

This means that the sum of arbitrary (at least 2 and at most $n$, not necessarily distinct) elements of $P$ modulo $p$, is not an element of $P$!

Note that $|P| = \frac{2p-1}{2n-1} - \frac{p-n}{2n-1} = \frac{p+n-1}{2n-1}$. Let $M = \{m_1, m_2, \dots, m_{2n^2-3n+2}\}$. For each $i = 1, 2, \dots, 2n^2-3n+2$, the numbers $\{m_i \cdot 1, m_i \cdot 2, \dots, m_i \cdot (p-1)\}$ form a permutation of $\{1, 2, \dots, (p-1)\}$ (mod $p$), i.e., exactly $\frac{p+n-1}{2n-1}$ of them are elements of $P$! Thus, there exists $k \in \{1, 2, \dots, p-1\}$, such that among the numbers $\{k \cdot m_1, k \cdot m_2, \dots, k \cdot m_{2n^2-3n+2}\}$ at least
$$\frac{(2n^2 - 3n + 2) \cdot \frac{p+n-1}{2n-1}}{p-1} > \frac{2n^2 - 3n + 2}{2n-1} > n-1$$
are elements of $P$ modulo $p$. It is easy to check, that the corresponding elements $\{m_j\}$ (at least $n$ in number) of $M$, satisfying $k \cdot m_j \in P$ (mod $p$), fulfill the problem statement. In particular, so does every their subset of exactly $n$ elements.