Example 2. There are $n$ points on a plane, where any three points can be covered by a circle of radius 1, but there are always three points that cannot be covered by any circle of radius less than 1. Find the minimum radius of a circle that can cover all $n$ points.
We prove that the radius of the smallest circle covering $n$ points is 1.

Since any three points can be covered by a circle with a radius of 1, the distance between any two points is no more than 2.

Thus, a circle with a radius of 2 centered at any one of the points can cover all the points. This implies that there exists a circle that can cover all the points, and therefore, there must also exist a smallest circle that can cover all the points. Let this circle be $\odot C$, with radius $r$. By the problem statement, $r \geqslant 1$.
We now prove that $r \leqslant 1$.
If there is only one point or no points on the circumference of $\odot C$ among the eight points, then there must be a circle with a radius smaller than $r$ that can cover all the points. $\odot C$ would not be the smallest circle covering all the points, which is impossible. Therefore, the circumference of $\odot C$ must have at least two points.

Assume that $\odot C$ has exactly two points, then these two points must be at the ends of a diameter, otherwise $\odot C$ can still be reduced. Since the distance between the two points is no more than 2, the radius $r$ of $\odot C$ is $\leqslant 1$.

If $\odot C$ has more points, $a_{1}, a_{2}, \cdots, a_{k}(3 \leqslant k \leqslant n)$, then these $k$ points cannot all be on a minor arc, otherwise $\odot C$ can still be reduced. This means that at least three points are on a major arc, and the triangle formed by these three points does not contain an obtuse angle. $\odot C$ is the smallest circle containing these three points. Since a circle with a radius of 1 that covers these three points can completely cover $\odot C$, it follows that $r \leqslant 1$.
Since $r \geqslant 1, r \leqslant 1$, thus $r=1$.