China Western Mathematical Olympiad

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 $a_k=x_1+\cdots +x_k$, $k=1,2,\dots ,2n$. Using Cauchy's Inequality, we have $$\begin{array}{rl}& (a_{n+1}+a_{n+2}+\cdots +a_{2n})-(a_1+a_2+\cdots +a_n)\\ & =n(x_1+\cdots +x_n)+n{x}_{n+1}+(n-1)x_{n+2}+\cdots +x_{2n}\\ & -[n{x}_1+(n-1)x_2+\cdots +x_n]\end{array}$$

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

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