Cesko-Slovacko-Poljsko

Let $|A|=m\ge 4n\sqrt{n}$ and let $S$ be the set of all segments with endpoints in $A$. Clearly, $|S|=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. The coordinates of every midpoint of a segment from $S$ are integer multiples of $1/2$. In the convex hull of $G$ there are less than $4n^2$ such points so there exists a point $B$ which is a midpoint of at least $(m/2)/(4n^2)$ segments from $S$. Let $P$ be the set of all segments from $S$ with their midpoints in $B$. Then $$|P|\ge \frac{\left(\genfrac{}{}{0ex}{}{m}{2}\right)}{4n^2}=\frac{m(m-1)}{8n^2}\ge \frac{4n\sqrt{n}(4n\sqrt{n}-1)}{8n^2}=\frac{16n^3-4n\sqrt{n}}{8n^2}=2n-\frac{1}{2\sqrt{n}}>2n-1.$$ and $|P|\ge 2n$.

Let us divide $P$ into disjoint families of segments which lay on the same line. Suppose that the number of such families is $k$ and in the $i$-th family we have $a_i$ segments, for $i=1,\dots ,k$. Every segment from $a_i$ segments of one family has its endpoints in $G$ and they have a common midpoint, so $a_i\le n/2$. Moreover every two segments from $P$ are the diagonals of a parallelogram iff they do not lay on the same line. Therefore for the number of different parallelograms with diagonals belonging to $P$ we have $$\sum _{1\le i<j\le k}{a}_i{a}_j=\frac{1}{2}\left({\left(\sum _{i=1}^k{a}_i\right)}^2-\sum _{i=1}^k{a}_i^2\right)\ge \frac{1}{2}\left({\left(\sum _{i=1}^k{a}_i\right)}^2-\sum _{i=1}^k{a}_i\cdot \frac{n}{2}\right)=\frac{1}{2}\left(|P{|}^2-|P|\cdot \frac{n}{2}\right)=\frac{1}{2}|P|\left(|P|-\frac{n}{2}\right)\ge n\left(2n-\frac{n}{2}\right)=\frac{3}{2}{n}^2>n^2.$$ Thus we have more than $n^2$ convex quadrangles (parallelograms) satisfying the given condition.