Argentine National Olympiad 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 $$|n-b_k|+|b_k-b_{k-1}|+\cdots +|b_1-1|\ge (n-b_k)+(b_k-b_{k-1})+\cdots +(b_1-1)=n-1$$ $$|n-c_l|+|c_l-c_{l-1}|+\cdots +|c_1-1|\ge (n-c_l)+(c_l-c_{l-1})+\cdots +(c_1-1)=n-1$$ 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 $S=|a_1-a_2|+|a_2-a_3|+\cdots +|a_{n-1}-a_n|+|a_n-a_1|\ge 2n-2$. 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 $b_k>b_{k-1}>\cdots >b_1$ and $c_l>c_{l-1}>\cdots >c_1$ (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 $\{2,\dots ,n-1\}$.