Points $A_1, A_2, \ldots, A_{2022}$ are chosen on a plane so that no three of them are collinear. Consider all angles $A_iA_jA_k$ for distinct points $A_i, A_j, A_k$. What largest possible number of these angles can be equal to $90^\circ$?

[i]Proposed by Anton Trygub[/i]
1. **Fixing a Point and Counting Right Angles:**
   - Fix a point \( A_j \) and count the number of right angles with \( A_j \) as a vertex.
   - Note that any two such right angles cannot share a segment, since then the other points of these angles would be collinear.
   - Each right angle takes two unique segments. Therefore, fixing \( A_j \) allows us at most \( \left\lfloor \frac{2021}{2} \right\rfloor = 1010 \) right angles with a vertex at \( A_j \).

2. **Calculating the Total Number of Right Angles:**
   - Since there are \( 2022 \) choices for \( A_j \), the total number of right angles is at most:
     \[
     1010 \cdot 2022 = 2{,}042{,}220
     \]

3. **Verification with a Regular Polygon:**
   - Consider a regular \( 2022 \)-gon. In such a polygon, there are \( 1011 \) pairs of diametrically opposite points.
   - For each pair of diametrically opposite points, there are \( 2020 \) remaining points that can serve as the vertex of the right angle.
   - Therefore, the number of right angles in this configuration is:
     \[
     1011 \cdot 2020 = 2{,}042{,}220
     \]

4. **Conclusion:**
   - The configuration of a regular \( 2022 \)-gon achieves the maximum number of right angles, confirming that the upper bound is tight.

\(\blacksquare\)

The final answer is \( \boxed{ 2{,}042{,}220 } \)