26124 ** There are $2 n$ points on a straight line, the distance between adjacent points is 1, a person starts jumping from the 1st point to other points, after $2 n$ jumps, he returns to the 1st point, these $2 n$ jumps cover all $2 n$ points, how to jump to make the total distance he jumps the farthest?
If we follow the path $1 \rightarrow 2 n \rightarrow 2 \rightarrow 2 n-1 \rightarrow 3 \rightarrow \cdots \rightarrow n+2 \rightarrow n \rightarrow n+1 \rightarrow 1$, then the total distance is $(2 n-1)+(2 n-2)+(2 n-1-2)+(2 n-1-3)+\cdots+[n+2-(n-1)]$ $+(n+2-n)+(n+1-n)+(n+1-1)=2 n-1+2 n-2+2 n-3+\cdots+3+$ $2+1+n=2 n^{2}$. We now prove that $2 n^{2}$ is the maximum distance.

In fact, let $a_{1}=1, a_{k}$ represent the point reached after the $(k-1)$-th jump $(2 \leqslant k \leqslant 2 n)$, and denote $a_{2 n+1}=$ $a_{1}$. Then the total distance $S=\sum_{i=1}^{2 n}\left|a_{i}-a_{i+1}\right|$. When the absolute value symbols are removed from $S$, each $a_{i}$ appears exactly twice, with $2 n$ instances as minuends and $2 n$ instances as subtrahends. The maximum value of $S$ is clearly achieved when the minuends are all $n+1-2 n$ and the subtrahends are all $1-n$. In this case, $S=2(n+1+n+2+\cdots+2 n)-$ $2(1+2+\cdots+n)=2 n^{2}$.
Therefore, the maximum value of $S$ is $2 n^{2}$, which means the maximum distance is $2 n^{2}$.