8. (Mathematics for Middle School, Issue 1, 2004 Olympiad Training Problem) Let $n$ be a given positive integer, and $n \geqslant 3$. For $n$ real numbers $x_{1}$, $x_{2}, \cdots, x_{n}$, denote the minimum value of $\left|x_{i}-x_{j}\right|(1 \leqslant i<j \leqslant n)$ as $m$. If $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, try to find the maximum value of $m$.
8. Let $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$, then
$$
x_{2}-x_{1} \geqslant m, x_{3}-x_{2} \geqslant m, \cdots, x_{n}-x_{n-1} \geqslant m \text {. }
$$

Thus, $x_{j}-x_{i} \geqslant(j-i) m(1 \leqslant i<j \leqslant n)$.
Hence $\sum_{1 \leqslant i<j \leqslant n}\left(x_{i}-x_{j}\right)^{2} \geqslant m^{2} \sum_{1 \leqslant i<j \leqslant n}(i-j)^{2}=m^{2} \sum_{k=1}^{n-1} \frac{k(k+1)(2 k+1)}{6}$
$$
\begin{array}{l}
=\frac{m^{2}}{6} \sum_{k=1}^{n-1}[2 k(k+1)(k+2)-3 k(k+1)]=\frac{m^{2}}{6} \sum_{k=1}^{n-1}\left(12 C_{k+2}^{3}-6 C_{k+1}^{2}\right) \\
=m^{2}\left(2 \sum_{k=1}^{n-1} C_{k+2}^{3}-\sum_{k=1}^{n-1} C_{k+1}^{2}\right)=m^{2}\left(2 C_{n+2}^{4}-C_{n+1}^{3}\right) \\
=\frac{1}{12} m^{2} n^{2}\left(n^{2}-1\right) .
\end{array}
$$

On the other hand, by $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, we get
$$
\sum_{1 \leqslant i<j \leqslant n}\left(x_{i}-x_{j}\right)^{2}=n-1-2 \sum_{1 \leqslant i \leqslant j \leqslant n} x_{i} x_{j}=n-\left(\sum_{k=1}^{n} x_{k}\right)^{2} \leqslant n \text {. }
$$

Therefore, $\frac{1}{12} m^{2} n^{2}\left(n^{2}-1\right) \leqslant n$.
Hence $m \leqslant \sqrt{\frac{12}{n\left(n^{2}-1\right)}}$.
Equality holds if and only if $x_{1}, x_{2}, \cdots, x_{n}$ form an arithmetic sequence, and $\sum_{k=1}^{n} x_{k}=0$.
Thus, $m_{\max }=\sqrt{\frac{12}{n\left(n^{2}-1\right)}}$.