II. (40 points) Let $n(n \geqslant 3)$ be a given natural number, and for $n$ given real numbers $a_{1}, a_{2}, \cdots, a_{n}$, denote the minimum value of $\left|a_{i}-a_{j}\right|(1 \leqslant i < j \leqslant n)$ as $m$. If $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$, find the maximum value of $m$.
Let's assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$.
On one hand,
$$
\begin{array}{l}
\sum_{1 \leqslant i < j \leqslant n}\left(a_{j}-a_{i}\right)^{2} \geqslant n \sum_{i=1}^{n-1}\left(a_{i+1}-a_{i}\right)^{2} \geqslant n(n-1) m^{2} .
\end{array}
$$
On the other hand, for $i < j$, we have
$$
\begin{array}{l}
a_{j}-a_{i}=\sum_{i=i}^{j-1}\left(a_{i+1}-a_{t}\right) \geqslant(j-i) m . \\
\text{Then } \sum_{1<i<j<n}\left(a_{j}-a_{i}\right)^{2} \geqslant m^{2} \sum_{1<i<j<n}(j-i)^{2} \\
=m^{2} \sum_{k=1}^{n-1}\left(1^{2}+2^{2}+\cdots+k^{2}\right) \\
=m^{2} \sum_{i=1}^{n-1} k^{2}(n-k) \\
=m^{2}\left[n \cdot \frac{n-1) n(2 n-1)}{6}-\frac{n^{2}(n-1)^{2}}{4}\right] \\
=\frac{m^{2}}{12} n^{2}\left(n^{2}-1\right) .
\end{array}
$$

Thus, $n \geqslant \frac{m^{2}}{12} n^{2}\left(n^{2}-1\right) \Rightarrow m \leqslant \sqrt{\frac{12}{n\left(n^{2}-1\right)}}$,
with equality holding if and only if $\sum_{i=1}^{n} a_{i}=0$, and $\left\{a_{i}\right\}$ forms an arithmetic sequence.
Therefore, the maximum value of $m$ is $\sqrt{\frac{12}{n\left(n^{2}-1\right)}}$.