10. 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 $\left(a_{n+1}+a_{n+2}+\cdots+a_{2 n}\right)$ $-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$.
10. When $n=1$, $\left(a_{2}-a_{1}\right)^{2}=1$, hence $a_{2}-a_{1}= \pm 1$, and the maximum value sought is 1.

For $n \geqslant 2$, let $x_{1}=a_{1}, x_{i+1}=a_{i+1}-a_{i}, i=1,2, \cdots, 2 n-1$, then
$$
\begin{array}{l}
\sum_{i=1}^{2 n} x_{i}^{2}=1, \text { and } a_{k}=x_{1}+x_{2}+\cdots+x_{k}, k=1,2, \cdots, 2 n . \text { By the Cauchy-Schwarz inequality, } \\
\left(a_{n+1}+a_{n+2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right) \\
=n\left(x_{1}+x_{2}+\cdots+x_{n}\right)+n x_{n+1}+(n-1) x_{n+2}+\cdots+x_{2 n}-\left[n x_{1}+(n-1) x_{2}+\cdots+x_{n}\right] \\
=x_{2}+2 x_{3}+\cdots+(n-1) x_{n}+n x_{n+1}+(n-1) x_{n+2}+\cdots+x_{2 n} \\
\leqslant\left[1^{2}+2^{2}+\cdots+(n-1)^{2}+n^{2}+(n-1)^{2}+\cdots+1\right]^{\frac{1}{2}}\left(x_{2}^{2}+x_{3}^{2}+\cdots+x_{2 n}^{2}\right)^{\frac{1}{2}} \\
=\left[n^{2}+2 \cdot \frac{1}{6} n(n-1)(2 n-1)\right]^{\frac{1}{2}} \\
=\sqrt{\frac{n\left(2 n^{2}+1\right)}{3}} .
\end{array}
$$

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

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 $\sqrt{\left.\frac{n\left(2 n^{2}\right.}{3}+1\right)}$.