, 2021

There are at least $4037$ points in $M$.
Since there are finitely many points in $S$, we can find a coordinate system such that all points in $S$ have distinct $x$-coordinates. Indeed, we just need to find a line which is not perpendicular to any line joining two of the points in $S$.

Now, let the $n = 2020$ points be $P_j = (x_j, y_j)$ where $j = 1, 2, \dots, n$. WLOG assume $x_1 < x_2 < \dots < x_n$. Since
$$\frac{x_1 + x_2}{2} < \frac{x_1 + x_3}{2} < \dots < \frac{x_1 + x_n}{2} < \frac{x_2 + x_n}{2} < \dots < \frac{x_{n-1} + x_n}{2},$$
the midpoints of $P_1P_2, P_1P_3, \dots, P_1P_n, P_2P_n, \dots, P_{n-1}P_n$ have pairwise different $x$-coordinates, and so they must be distinct. This shows $|M| \ge 2n - 3 = 4037$.

The case $4037$ is attainable. For example, we choose the points $0, 2, 4, 6, \dots, 2n-2$ on the real number line. Then the midpoints can only be $1, 2, 3, \dots, 2n-3$. So there are exactly $2n-3 = 4037$ midpoints.