Problem 1667

First, for $n=1$, we have $(a_2 - a_1)^2 = 1$, $a_2 - a_1 = \pm 1$. Then the maximum value of $a_2 - a_1$ is $1$.

Secondly, for $n \ge 2$, let $x_1 = a_1$, $x_{i+1} = a_{i+1} - a_i$, $i = 1, 2, \dots, 2n-1$. Then $\sum_{i=2}^{2n} x_i^2 = 1$, and $a_k = x_1 + \dots + x_k$, $k = 1, 2, \dots, 2n$. Using Cauchy's Inequality, we have
$$\begin{aligned} & (a_{n+1} + a_{n+2} + \dots + a_{2n}) - (a_1 + a_2 + \dots + a_n) \\ &= n(x_1 + \dots + x_n) + n x_{n+1} + (n-1)x_{n+2} + \dots + x_{2n} \\ &\quad - [n x_1 + (n-1)x_2 + \dots + x_n] \end{aligned}$$

$$\begin{align*} &= x_2 + 2x_3 + \cdots + (n-1)x_n + n x_{n+1} + (n-1)x_{n+2} + \cdots + x_{2n} \\ &\le [1^2 + 2^2 + \cdots + (n-1)^2 + n^2 + (n-1)^2 + \cdots + 1^2]^{\frac{1}{2}} (x_2^2 + x_3^2 + \cdots + x_{2n}^2)^{\frac{1}{2}} \\ &= \left[ n^2 + 2 \times \frac{1}{6}(n-1)n(2n-1) \right]^{\frac{1}{2}} = \sqrt{\frac{n(2n^2+1)}{3}}. \end{align*}$$

The equality holds when
$$\begin{gather*} a_k = \frac{\sqrt{3k(k-1)}}{2\sqrt{n(2n^2+1)}}, \quad a_{n+k} = \frac{\sqrt{3}[2n^2 - (n-k)(n-k+1)]}{2\sqrt{n(2n^2+1)}}, \\ k = 1, 2, \dots, n. \end{gather*}$$
So the maximum value of $(a_{n+1} + a_{n+2} + \cdots + a_{2n}) - (a_1 + a_2 + \cdots + a_n)$ is $\sqrt{\frac{n(2n^2+1)}{3}}$.