Consider all line segments formed by the points and partition them into groups of parallel segments. As no 3 points are collinear, any segment between any two of the points is a side of at most 2 parallelograms of unit area. On the other hand, consider all groups of segments. In each group, the topmost and bottommost segment (if we orient the segment to be horizontal) is a part of at most one parallelogram of unit area, and if there is only one segment it is not a part of any parallelogram. There are at least $n$ groups, as for three consecutive points $X,Y,Z$ on the convex hull of the points there are all lines emanating from $Y$ and $XZ$, no two of which are parallel. Hence, we subtract off $2n$, leaving $2\binom n2-2n=n^2-3n$ segment -- parallelogram pairs. Each parallelogram is a part of 4 of these, yielding the result.

Here's a bound that is slightly better, for sufficiently large $n$: Suppose there exists total $k$ such parallelograms. We note some propositions whose proofs are not hard: Each segment can be a side of at most two such parallelograms. Each segment on convex hull can be a side of at most one such parallelograms, and can't be diagonal of any of such parallelogram. Diagonal of any of such parallelogram can't be a side of any other such parallelogram. Each segment can be a diagonal of at most two such parallelograms. So, $k$ parallelograms, each has two diagonals, contributes at least $k$ distinct diagonals. Now, we count the number of pairing $(s,\mathcal{Q})$ where $s$ is a side of $\mathcal{Q}$, which is one of such $k$ parallelogram. Each of $k$ parallelograms has $4$ sides, so there're $4k$ such pairings. On the other hand, $s$ must be one of non-diagonal sides. Suppose the convex hull is a $c$-gon. So, the number of such pairing is at most $2\cdot \left( \binom{n}{2}-k-c\right) +c$. Hence, $$2\cdot \left( \binom{n}{2}-k\right) -c \geqslant 4k\implies \frac{n^2-n-c}{6}\geqslant k\implies \left\lfloor \frac{n^2-n-c}{6}\right\rfloor \geqslant k.$$Since $ \left\lfloor \frac{n^2-n-c}{6}\right\rfloor \leqslant \left\lfloor \frac{n^2-n-3}{6}\right\rfloor \leqslant \frac{n^2-3n}{4}$ is true for all $n\geqslant 4$, we are done.

This was the Balkan BMO Shortlist 2017 C3 problem.

This was created by me (Silouanos Brazitikos) and the proposed bound was obtained by Demetres Christophides (aka Demetres). The order $O(n^2)$ is optimal.

I wonder if we can prove it without "no 3 of them are conlinear".