We run a greedy algorithm to find an antiparallelogram set $X$ contained in $S$. Set $X = \emptyset$ 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:
$$2023 - k \le 3 \binom{k}{3}.$$
However $3\binom{16}{3} + 16 = 1696 < 2023$. Thus, $k \ge 17$, and we are done.