24. Points $A_{1}, \cdots, A_{5}$ are on a sphere with radius 1, what is the maximum value of $\min _{1 \leqslant \mathrm{i}, \mathrm{j} \leqslant 5} A_{1} A_{\mathrm{j}}$? Determine all cases where the maximum value is achieved.
24. We can use the size of $\angle A_{1} O A_{1}$ to replace the distance $A_{1} A_{\mathrm{j}}$, where $O$ is the center of the sphere.
There exists a set of points $A_{1}, A_{2}, \cdots, A_{5}$, such that
$$
\mathrm{mn}_{1 \leqslant \mathrm{i}, \mathrm{j}} \leqslant A_{1} O A_{\mathrm{j}} \leqslant \frac{\pi}{2} .
$$

For example, take 5 vertices of a regular octahedron.
Now let $A_{1}, A_{2}, \cdots, A_{5}$ be any set of points that satisfy (24.1) - we assert that there must be two antipodal points.

First, set $A_{5}$ as the South Pole. In this case, $A_{1}, A_{2}, A_{\mathrm{j}}, A_{4}$ are all in the Northern Hemisphere (including the equator). If there are no antipodal points, then the North Pole has no points. If a quadrant in the Northern Hemisphere, bounded by a quarter of the equator and two mutually perpendicular meridians, contains two $A_{\mathrm{t}}$, then due to (24.1), one must be at the "corner" (i.e., the intersection of the equator and a meridian), and the other must be on the opposite meridian. Thus, the difference in longitude between any two of $A_{1}, \cdots, A_{4}$ is less than $\frac{\pi}{2}$. Therefore, they lie on the four meridians that divide the Northern Hemisphere into four quadrants. Finally, if one point is not on the equator, then its two "neighbors" must be on the equator (each in a quadrant's "corner"), and these two points are antipodal.

Since there are always two antipodal points $A_{1}, A_{2}$, and the third point $A_{3}$ cannot simultaneously satisfy $\angle A_{1} O A_{3} > \frac{\pi}{2}$ and $\angle A_{2} O A_{3} > \frac{\pi}{2}$, the strict inequality in (24.1) cannot hold. Thus, the maximum value of $\min \angle A_{1} O A_{j}$ is $\frac{\pi}{2}$. When the equality holds, let $A_{1}, A_{2}$ be the two poles, then $A_{3}, A_{4}, A_{5}$ are all on the equator, and the angles between $O A_{3}, O A_{4}$, and $O A_{5}$ are not less than $\frac{\pi}{2}$.
The maximum value of $\min A_{1} A_{\mathrm{j}}$ is $\sqrt{2}$, which is achieved in the above situation.