** A set of good points on a plane, where any three points are vertices of a right triangle. Determine the maximum number of points this set can contain.
Let such a point set contain at most $n$ points. We first make an (as small as possible) upper bound estimate for $n$.

From the condition, we know that any three points in the point set lie on the circumference of a circle determined by the longest segment among these three points as its diameter. Therefore, if we take the two points in the point set that are farthest apart, denoted as $A$ and $B$, then all points in the point set lie on the circumference of the circle with $A B$ as its diameter. Moreover, it is easy to see that at most one point (different from $A$ and $B$) can lie on the semicircle with $A B$ as its diameter, for if there were two points $C$ and $D$, then $A$, $C$, and $D$ could not be the vertices of a right triangle. Thus, $n \leqslant 4$.
On the other hand, the set of the four vertices of a square clearly meets the requirement, so the required $n$ is 4.