For $101 n$ as an integer not less than 2, determine the largest constant $C(n)$ such that
$$
\sum_{1 \leqslant i<j \leqslant n}\left(x_{j}-x_{i}\right)^{2} \geqslant C(n) \cdot \min \left(x_{i+1}-x_{i}\right)^{2}
$$

holds for all real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}<x_{2}<\cdots<x_{n}$.
$$
\begin{array}{l}
\text { Solution: } \because x_{j}-x_{i}=\left(x_{j}-x_{j-1}\right)+\left(x_{j-1}-x_{j-2}\right) \\
+\cdots+\left(x_{i+1}-x_{i}\right) \\
\geqslant(j-i) \cdot \min \left(x_{i+1}-x_{i}\right),(j>i) \\
\therefore \frac{\sum_{1 \leqslant i<j \leqslant n}\left(x_{j}-x_{r}\right)^{2}}{\min \left(x_{i+1}-x_{i}\right)^{2}} \geqslant \sum_{1 \leqslant i<j \leqslant n}(j-i)^{2} \\
=\sum_{i=1}^{n-1}(n-i) i^{2}=n \sum_{i=1}^{n-1} i^{2}-\sum_{i=1}^{n} i^{3} \\
=\frac{n(n-1) n[2(n-1)+1]}{6}-\left[\frac{(n-1) n}{2}\right]^{2} \\
=\frac{1}{12} n^{2}\left(n^{2}-1\right) \text {. } \\
\end{array}
$$

Therefore, $C(n)=\frac{1}{12} n^{2}\left(n^{2}-1\right)$.