Rioplatense Mathematical Olympiad

We run a greedy algorithm to find an antiparallelogram set $X$ contained in $S$. Set $X=\varnothing$ to start. In each step, verify if there are points in $S\setminus X$ that can be incorporated to $X$ while keeping it antiparallelogram. If so, choose any one of those points, add it to $X$, and repeat. Else, the algorithm stops. At the end we have an antiparallelogram set $X$ with $k$ points such that for any point $p\in S\setminus X$, $X\cup \{p\}$ is not antiparallelogram. In other words, for any $p\in S\setminus X$, there are $q,r,s\in X$ such that $p,q,r,s$ are the vertices of a parallelogram. But notice that for any $q,r,s$ there are exactly $3$ such $p$'s, so the following inequality must hold: $$2023-k\le 3\left(\genfrac{}{}{0ex}{}{k}{3}\right).$$ However $3\left(\genfrac{}{}{0ex}{}{16}{3}\right)+16=1696<2023$. Thus, $k\ge 17$, and we are done.