The 100th problem: Given $n(n \geq 2)$ points $A_{1}, A_{2}, \ldots,$ $A_{n}$ on the sphere $\left\{(x, y, z) \mid x^{2}+y^{2}+z^{2}=1\right\}$, find the maximum possible value of $\sum_{1 \leq \mathrm{i}<\mathrm{j} \leq \mathrm{n}}\left|A_{\mathrm{i}} \mathrm{A}_{\mathrm{j}}\right|^{2}$.
Question 100, Solution: Let the center of the sphere be 0, and define the vector $\alpha_{i}=\overrightarrow{0 A_{i}}(1 \leq \mathrm{i} \leq \mathrm{n})$, then $\left|\alpha_{i}\right|^{2}=1$. According to the Lagrange identity, we have:
$$
\sum_{1 \leq i<j \leq n}\left|A_{i} A_{j}\right|^{2}=\sum_{1 \leq i<j \leq n}\left(\alpha_{j}-\alpha_{i}\right)^{2}=n\left(\sum_{i=1}^{n}\left|\alpha_{i}\right|^{2}\right)-\left|\sum_{i=1}^{n} \alpha_{i}\right|^{2} \leq n\left(\sum_{i=1}^{n}\left|\alpha_{i}\right|^{2}\right)=n^{2}
$$

Take $\alpha_{1}=\left(\cos \frac{2 i \pi}{n}, \sin \frac{2 i \pi}{n}, 0\right)(1 \leq i \leq n)$, then the equality holds.