## Solution.

We can select at most 1012 points.

First, let's prove that there exists a selection of 1012 points among which no four points are vertices of a rectangle. Choose the points \( A_{1}, A_{2}, \ldots, A_{1012} \).

Assume that among these 1012 points, there are points \( A, B, C \), and \( D \) such that they form a rectangle \( ABCD \). The circle circumscribed around this rectangle is the same circle circumscribed around the regular 2022-gon \( A_{1} A_{2} \ldots A_{2022} \). Let \( S \) be the center of this circle. The center of the circle circumscribed around a rectangle is the intersection of its diagonals, which bisects the diagonals. Thus, point \( S \) is the midpoint of diagonals \( \overline{AC} \) and \( \overline{BD} \), which is possible only if the pairs of points \( A \) and \( C \), and \( B \) and \( D \) are centrally symmetric with respect to \( S \). On the other hand, among the points \( A_{1}, A_{2}, \ldots, A_{1012} \), there are no two pairs of such points (only one such pair exists, consisting of \( A_{1} \) and \( A_{1012} \)). Therefore, among these 1012 points, no four points form a rectangle.

Now, let's prove that among any 1013 points, there exist four points that are vertices of a rectangle.

Consider 1011 pairs of points \( \left\{A_{1}, A_{1012}\right\}, \left\{A_{2}, A_{1013}\right\}, \ldots, \left\{A_{1011}, A_{2022}\right\} \).

2 points

By selecting 1013 points from these 1011 pairs, there will be at least two pairs of points in which we have selected all four points of those two pairs.

Let's prove that these four points form the vertices of a rectangle.

The center of the circle circumscribed around the regular 2022-gon \( S \) is the midpoint of the segments formed by pairs of centrally symmetric points. Additionally, point \( S \) lies on both segments formed by these pairs. Therefore, \( S \) is necessarily the intersection of the diagonals of the quadrilateral formed by the four selected points. Point \( S \) bisects these diagonals, and they are of equal length (equal to the diameter of the circumscribed circle). Any quadrilateral whose diagonals are bisected by their intersection and are of equal length is necessarily a rectangle. Therefore, these four points are indeed the vertices of a rectangle.

Thus, the statement is proven: we can select at most 1012 vertices of a 2022-gon such that no four of the selected points form the vertices of a rectangle.

Note: In the above solution, we have proven the following statement: four points of a regular 2022-gon form a rectangle if and only if among them there are two pairs of centrally symmetric points with respect to \( S \). The proof of this statement in the above solution corresponds to the parts of the solution that are worth the third and fourth points, and the last three points. The part of the solution where this statement is stated is worth an additional 1 point if neither of its directions is proven in the rest of the solution.

The justification that among the points \( A_{1}, A_{2}, \ldots, A_{1012} \) no four points form the vertices of a rectangle can also be derived differently. In the following, we provide another proof. All 1012 points (and any four of them) lie on one half of the circle circumscribed around the regular 2022-gon. Consider the vertex with the smallest index among any four. The internal angle of the quadrilateral at this vertex is necessarily acute, because it is an inscribed angle over a chord whose endpoints lie on the same semicircle (certainly this chord is not a diameter of the circle), so this quadrilateral cannot be a rectangle.
## Task A-4.4.

Given is a regular 2022-gon $A_{1} A_{2} \ldots A_{2022}$. What is the maximum number of vertices of the regular 2022-gon that we can select so that no four of the selected points form the vertices of a rectangle?