The answer is $\boxed{2{,}042{,}220}$. Fix $A_j$ and count the number of right angles with it 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. It follows that, since each right angle takes two unique segments, fixing $A_j$ allows us at most $\left\lfloor\frac{2021}{2}\right\rfloor = 1010$ right angles with a vertex at $A_j$. Since there are $2022$ choices for $A_j$, the answer is at most $1010 \cdot 2022 = 2{,}042{,}220$. Indeed, this is tight if we take a regular $2022$-gon: there are $1011$ pairs of diametrically opposite points, and then $2020$ points remaining to be the vertex, so there are $1011 \cdot 2020 = 2{,}042{,}220$ right angles, as desired.

Posted by Michael Ren in 2016. Nice result though.