$7 \cdot 1$ Given $n$ points on a plane, any 3 of which are the 3 vertices of a right-angled triangle, find the maximum value of $n$.

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[Solution] Clearly, any 3 points out of the 4 vertices of a rectangle form the 3 vertices of a right-angled triangle. Therefore, the maximum value of the required $n$ is no less than 4.

If there are 5 points that satisfy the condition, let the two points with the maximum distance among the 5 points be $A$ and $B$. Then the other 3 points must lie on the circle with $AB$ as its diameter. By the pigeonhole principle, among these 3 points, there must be two points in the same semicircle, say $C$ and $D$. Thus, $\triangle ACD$ cannot be a right-angled triangle, which is a contradiction. In summary, the maximum value of the required $n$ is 4.