, 2016

First we clarify the meaning of the given equality. Let $a_1, a_2, ..., a_n$ be an arbitrary circular arrangement of $1, 2, ..., n$, $n \ge 3$, in clockwise direction. The extremal numbers $1$ and $n$ separate the remaining numbers into two groups. For convenience denote them by $b_1, ..., b_k$ and $c_1, ..., c_l$, arranged as shown in the figure. Here $k + l = n - 2$; one of $k$ and $l$ can be zero. We have
$$|n - b_k| + |b_k - b_{k-1}| + \dots + |b_1 - 1| \ge (n - b_k) + (b_k - b_{k-1}) + \dots + (b_1 - 1) = n - 1$$
$$|n - c_l| + |c_l - c_{l-1}| + \dots + |c_1 - 1| \ge (n - c_l) + (c_l - c_{l-1}) + \dots + (c_1 - 1) = n - 1$$
The absolute values in the two left hand sides are
$$|a_1 - a_2|, |a_2 - a_3|, \dots, |a_{n-1} - a_n|, |a_n - a_1|$$
Adding up gives $S = |a_1 - a_2| + |a_2 - a_3| + \dots + |a_{n-1} - a_n| + |a_n - a_1| \ge 2n - 2$. For any circular arrangement $a_1, a_2, ..., a_n$ of $1, 2, ..., n$.
We are interested in the equality case. Clearly $S = 2n - 2$ if and only if $b_k > b_{k-1} > \dots > b_1$ and $c_l > c_{l-1} > \dots > c_1$ (because $n > b_k, b_1 > 1$ and $n > c_l, c_1 > 1$ hold trivially).
Now we show that there is a bijection between our admissible circular arrangements, the ones with $S = 2n - 2$, and the subsets of $\{2, \dots, n-1\}$. Let $B$ be any subset of $\{2, \dots, n-1\}$, including the empty one. Construct a circular arrangement of $1, 2, ..., n$ in a way suggested by the previous reasoning. Start with $1$, proceed in clockwise direction along the circle by placing the elements of $B$ in increasing order, place $n$ after them and finish with the remaining elements of $\{2, \dots, n-1\}$ in decreasing order. By the above the obtained circular arrangement is admissible, and clearly different subsets $B$ of $\{2, \dots, n-1\}$ give rise to different arrangements. The bijection shows that there are $2^{n-2}$ admissible circular arrangements, as many as the subsets of $\{2, \dots, n-1\}$.