Example 13 (2003 China Western Mathematical Olympiad) Let $2 n$ real numbers $a_{1}, a_{2}, \cdots, a_{2 n}$ satisfy the condition $\sum_{i=1}^{2 n+1}\left(a_{i+1}-a_{i}\right)^{2}=1$, find the maximum value of $\left(a_{n+1}+a_{n+2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$.
Proof: First, when $n=1$, $\left(\dot{a}_{2}-a_{1}\right)^{2}=1$, so $a_{2}-a_{1}= \pm 1$, and the maximum value sought is 1.
When $n \geqslant 2$, let $x_{1}=a_{1}, x_{i+1}=a_{i+1}-a_{i}, i=1,2,3, \cdots, 2 n-1$, then $\sum_{i=2}^{2 \pi} x_{i}^{2}=1$, and $a_{k}=$ $x_{1}+x_{2}+\cdots+x_{k}, k=1,2, \cdots, 2 n$.
Therefore, by the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
& \left(a_{n+1}+a_{n+2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right) \\
= & n\left(x_{1}+x_{2}+\cdots+x_{n}\right)+n x_{n+1}+(n-1) x_{n+2}+\cdots+x_{2 n}-\left[n x_{1}+(n-1) x_{2}+\cdots+x_{n}\right] \\
= & x_{2}+2 x_{3}+\cdots+(n-1) x_{n}+n x_{n+1}+(n-1) x_{n+2}+\cdots+x_{2 n} \\
\leqslant & {\left[1^{2}+2^{2}+\cdots+(n-1)^{2}+n^{2}+(n-1)^{2}+\cdots+1^{2}\right]^{+}\left(x_{2}^{2}+x_{3}^{2}+\cdots+x_{2 n}^{2}\right)^{+} } \\
= & {\left[n^{2}+2 \times \frac{1}{6}(n-1) n(2 n-1)\right]^{+}=\sqrt{\frac{n\left(2 n^{2}+1\right)}{3}} . }
\end{aligned}
$$

When $a_{k}=\frac{\sqrt{3} k(k-1)}{2 \sqrt{n\left(2 n^{2}+1\right)}}, k=1,2, \cdots, n, a_{n+k}=\frac{\sqrt{3}\left[2 n^{2}-(n-k)(n-k+1)\right]}{2 \sqrt{n\left(2 n^{2}+1\right)}}, k=1$, $\cdots, n$ the above equality holds.

Therefore, the maximum value of $\left(a_{n+1}+a_{n+2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$ is $\sqrt{\frac{n\left(2 n^{2}+1\right)}{3}}$.