$n$ points are on a plane, with no three collinear. What is the maximum possible number, in terms of $n$, of non-overlapping squares we can draw by drawing segments between pairs of these points in such a way that no two segments intersect in the interior?
If the non-collinearity condition is not there, then the maximum is (close to) $n$, obtained by drawing a square grid with side roughly $\sqrt{n}$. This is also (close to) optimal, because each point can be part of at most four squares, and each square contains four points. With the collinearity condition, each point can be part of at most three squares, so the maximum is at most around $3n/4$.