Question 44: Given 2015 points on a plane, the distance between any two points is unique. For each point, the 22 closest points to it are called its "neighbors". How many points can a given point be a neighbor to at most?
The 44th question:
Solution: First, we prove a lemma: If there are 2015 points in the plane with pairwise distinct distances, for each point, the point closest to it is called its nearest neighbor. Then, a given point can be the nearest neighbor of at most 5 points.

Proof of the lemma: For any point \( P \), if \( P \) is the nearest neighbor of \( A_{1} \) and \( A_{2} \), then \( d(A_{1}, P) \) and \( d(A_{2}, P) \) form an angle of \( 60^{\circ} \). Therefore, point \( P \) can be the nearest neighbor of at most 5 points. The lemma is proved!

Now, let's return to the original problem.
Let \( P \) be the neighbor of \( n \) points \( A_{1}, A_{2}, \ldots, A_{n} \).
Let \( A_{i} (i \in \{1, 2, \ldots, n\}) \) and \( P \) have the maximum distance. Denote \( A_{i} \) as \( F_{1} \), and remove all points within the circle centered at \( F_{1} \) with radius \( d(P, F_{1}) \) (including \( A_{i} \), but not including point \( P \)).

Among the remaining points, let \( A_{j} (j \in \{1, 2, \ldots, n\}) \) and \( P \) have the maximum distance. Denote \( A_{j} \) as \( F_{2} \), and remove all points within the circle centered at \( F_{2} \) with radius \( d(P, F_{2}) \).
Continue selecting new points \( F_{3}, F_{4}, \ldots, F_{t} \) that are farthest from point \( P \) until all points \( A_{1}, A_{2}, \ldots, A_{n} \) are removed.
According to the above rules, for any \( i \) and \( j \), \( d(P, F_{j}) > d(P, F_{i}) \), and \( d(F_{i}, F_{j}) > d(P, F_{i}) \).

Thus, for \( F_{1}, F_{2}, \ldots, F_{t} \), \( P \) is the nearest neighbor of each \( F_{i} (i \in \{1, 2, \ldots, t\}) \). According to the lemma, \( t \leq 5 \). Since each time at most 22 points are removed, a total of at most \( 5 \times 22 = 110 \) points can be removed, so \( n \leq 110 \).

On the other hand, consider a regular pentagon centered at \( P \), with 22 points around each vertex that are sufficiently close to each other, and each of these points has \( P \) as its nearest neighbor. Additionally, take 1904 other points that are sufficiently far from these 111 points. In this case, point \( P \) is the neighbor of exactly 110 points.
In summary, a point can be the neighbor of at most 110 points.
Note: Generally, for each point, the \( k \) closest points are called its "neighbors", then each point can be the neighbor of at most \( 5k \) points.