Problem 1527

1. **Define the set $G$ and subset $A$:**
- The set $G$ consists of $n^2$ points in the plane:
$$G = \{(x, y) \mid x, y \in \{1, 2, \ldots, n\}\}$$
- The subset $A$ is a subset of $G$ with at least $4n\sqrt{n}$ elements.

2. Objective:
- We need to prove that there are at least $n^2$ convex quadrilaterals whose vertices are in $A$ and all of whose diagonals pass through a fixed point.

3. Finding line segments with a common midpoint:
- Consider the number of possible line segments with endpoints in $A$. The number of such segments is given by:
$$\binom{|A|}{2} \ge \binom{4n\sqrt{n}}{2}$$
- Using the binomial coefficient approximation for large $n$:
$$\binom{4n\sqrt{n}}{2} = \frac{4n\sqrt{n}(4n\sqrt{n} - 1)}{2} \approx 8n^3 - 2n^{3/2}$$

4. Possible midpoints:
- Each line segment has a midpoint, and the number of possible midpoints is less than $4n^2$ because there are $n^2$ points in $G$ and each point can be a midpoint of a segment.

5. Applying the pigeonhole principle:
- Since there are $8n^3 - 2n^{3/2}$ possible segments and fewer than $4n^2$ possible midpoints, by the pigeonhole principle, there must be at least one midpoint that is the midpoint of at least:
$$\frac{8n^3 - 2n^{3/2}}{4n^2} = 2n - \frac{n^{3/2}}{2n^2} = 2n - \frac{1}{2n}$$
line segments. For large $n$, this is approximately $2n$ line segments.

6. Forming parallelograms:
- We have found at least $2n$ line segments with a common midpoint. Any pair of these segments can form the diagonals of a parallelogram. The number of ways to choose 2 segments from $2n$ segments is:
$$\binom{2n}{2} = \frac{2n(2n - 1)}{2} = 2n^2 - n$$
- Since $2n^2 - n \ge n^2$ for $n \ge 1$, we have at least $n^2$ parallelograms.

7. Conclusion:
- Therefore, there are at least $n^2$ convex quadrilaterals (parallelograms) whose vertices are in $A$ and all of whose diagonals pass through a fixed point.

$\blacksquare$