2) 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 $\left(a_{n-1}+a_{n-2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$. (Leng Gangsong provided the problem)
Given that the maximum value desired is 1.

When $n \geq 2$, let $x_{1}=a_{1}, x_{i-1}=a_{i+1}-a_{i}, i=1,2, \cdots, 2 n-1$, then $\sum^{2 n} x_{i}^{2}=1$, 11
$$
a_{k}=x_{1}+\cdots+x_{k} \cdot k-1,2, \cdots, 2 n
$$

Therefore, by the Cauchy-Schwarz inequality:
$$
\begin{aligned}
& \left(a_{n+1}+a_{n+2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right) \\
= & n\left(x_{1}+\cdots+x_{n}\right)+n x_{n+1}+(n-1) x_{n-2}+\cdots+x_{2 n}- \\
& {\left[n x_{1}+(n-1) x_{2}+\cdots+x_{n}\right\rfloor } \\
= & x_{2}+2 x_{3}+\cdots+(n-1) x_{n}+n x_{n-1}+(n-1) x_{n-2}+\cdots+x_{2 n} \\
\leqslant & {\left[1^{2}+2^{2}+\cdots+(n-1) 2+n^{2}+(n-1)^{2}+\cdots+1^{2}\right]^{\frac{1}{2}} } \\
& \left(x_{2}^{3}+x_{3}^{2}+\cdots+x_{2}^{2}\right)^{\frac{1}{2}}
\end{aligned}
$$
$$
=\left[n^{2}+2 \times \frac{1}{6}(n-1) n(2(n-1)+1)\right]^{\frac{1}{2}}=\sqrt{\frac{n\left(2 n^{2}+1\right)}{3}},
$$
$\quad a_{k}=\frac{\sqrt{3} k(k-1)}{2 \sqrt{n\left(2 n^{2}+1\right)}} \cdot k=1, \cdots, n$,
$$
u_{n+k}=\frac{\sqrt{3}\left[2 n^{2}-(n-k)(n-k+1)\right]}{2 \sqrt{n}\left(2 n^{2}+1\right)}, k=1, \cdots, n
$$

When the above inequality holds with equality.
Thus, 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 $\sqrt{\frac{n\left(2 n^{2}+1\right)}{3}}$.