In fact, there are $n^2$ parallelograms all of whose diagonals pass through a given point. We will find $2n$ line segments with endpoints in $A$ and a common midpoint. There are ${4n\sqrt{n} \choose 2} \ge 8n^3-2n^{3/2}$ possible segments and less than $4n^2$ possible midpoints. Since $2n^{3/2}< 4n^2$, pigeonhole finds the $2n$ line segments for us. We can choose any pair of the segments to be diagonals for a parallelogram, yielding ${2n \choose 2}\ge n^2$ parallelograms a desired.

The above argument is not true. Not all pairs of those segments will form a parallelogram. Some of them may coincide with another. Like the segment (-1,0)-(1,0) coincides with (-2,0)-(2,0).

oneplusone wrote: The above argument is not true. Not all pairs of those segments will form a parallelogram. Some of them may coincide with another. Like the segment (-1,0)-(1,0) coincides with (-2,0)-(2,0). Actually at most $ \frac{n}{2} $ segments coincide, so we can get at least $ \binom{2n}{2}-4\binom{\frac{n}{2}}{2}\geq n^2 $