Baltic Way 1990.1. Numbers $1, \dots, n$ are written around a circle in some order. Determine the smallest possible sum of the absolute differences of adjacent numbers. Solution. The smallest possible sum is $2n-2$, achieved when the numbers are written in the order $1, \dots, n$. On the other hand, let the numbers are written in the order $a_1, \dots, a_n$. Take all subscripts modulo $n$, just note that since the $a_k$'s are pairwise distinct, there must exists indices $i$ and $j$ such that $|a_i-a_j|\ge n-1$, hence \[\sum\limits_{k=1}^n|a_k-a_{k+1}|=\sum\limits_{k=i}^{j-1}|a_k-a_{k+1}|+\sum\limits_{k=j}^{i-1}|a_k-a_{k+1}|\ge\left|\sum\limits_{k=i}^{j-1}(a_k-a_{k+1})\right|+\left|\sum\limits_{k=j}^{i-1}(a_k-a_{k+1})\right|=|a_i-a_j|+|a_j-a_i|\ge 2n-2.\]

Is it possible to find the maximum sum, too?