Isn’t this well known?

The answer is $4037$. Indeed, suppose $$S=\{(1,0),(2,0),...,(2020,0)\}$$then $$M=\{(\frac{3}{2},0),...,(\frac{2020+2019}{2},0)\}$$so $|M|=4037$ is attainable. Now we will prove the bound. Order the points in $S$ by $A_1,A_2,...,A_{2020}$, such that if $A_i=(x_i,y_i)$ in the Cartesian plane then $x_1\leq x_2\leq ...\leq x_{2020}$ and that if $x_i=x_{i+1}$ then $y_i<y_{i+1}$ Let $M_{i,j}$ denote the midpoint of $A_i$ and $A_j$, then $$\frac{a_1+a_2}{2}\leq \frac{a_1+a_3}{2}\leq...\leq\frac{a_1+a_{2020}}{2}\leq\frac{a_2+a_{2020}}{2}\leq...\leq \frac{a_{2019}+a_{2020}}{2}$$From this one can easily show that $M_{1,2},...,M_{1,2020},M_{2,2020},...,M_{2019,2020}$ are $4037$ distinct elements of $M$, this completes the proof. Quote: Isn’t this well known? Indeed it is, this is a fairly common trick in combinatorics problems.

mathaddiction wrote: The answer is $4037$. Indeed, suppose $$S=\{(1,0),(2,0),...,(2020,0)\}$$then $$M=\{(\frac{3}{2},0),...,(\frac{2020+2019}{2},0)\}$$so $|M|=4037$ is attainable. Now we will prove the bound. Order the points in $S$ by $A_1,A_2,...,A_{2020}$, such that if $A_i=(x_i,y_i)$ in the Cartesian plane then $x_1\leq x_2\leq ...\leq x_{2020}$ and that if $x_i=x_{i+1}$ then $y_i<y_{i+1}$ Let $M_{i,j}$ denote the midpoint of $A_i$ and $A_j$, then $$\frac{a_1+a_2}{2}\leq \frac{a_1+a_3}{2}\leq...\leq\frac{a_1+a_{2020}}{2}\leq\frac{a_2+a_{2020}}{2}\leq...\leq \frac{a_{2019}+a_{2020}}{2}$$From this one can easily show that $M_{1,2},...,M_{1,2020},M_{2,2020},...,M_{2019,2020}$ are $4037$ distinct elements of $M$, this completes the proof. Quote: Isn’t this well known? Indeed it is, this is a fairly common trick in combinatorics problems. Shouldn't it be 4039 after this argument ?

Singapore MO Senior Round 2 2017 https://artofproblemsolving.com/community/c1068820h2042481p14456003 But according to memory even the Singapore one was stolen from some old Russian competition.