The 37th Korean Mathematical Olympiad Final Round

The answer is $n^2$ if $n$ is even, and $n^2-5$ if $n$ is odd. It is easy when $n$ is even. Let $k=\frac{n+1}{2}$. Then, $$\begin{array}{rl}\sum _{i=1}^n|a_i-a_{i+1}+a_{i+2}-a_{i+3}|& \le \sum _{i=1}^n(|k-a_i|+|k-a_{i+1}|+|k-a_{i+2}|+|k-a_{i+3}|)\\ & =4\sum _{i=1}^n|k-a_i|=n^2,\end{array}$$ and the equality holds when $a_{2i-1}=i$ and $a_{2i}=\frac{n}{2}+1$ for $i=1,2,\dots ,\frac{n}{2}$. So, the maximum of ${\sum }_{i=1}^n|a_i-a_{i+1}+a_{i+2}-a_{i+3}|$ is $n^2$.

Let $n$ be odd. Without loss of generality, let $a_1-a_2>0$. If $a_i-a_{i+1}>0$ for all $i=1,2,\dots ,n$, then $$0=(a_1-a_2)+(a_2-a_3)+\cdots +(a_n-a_1)>0,$$ a contradiction. So, there exists $1<x\le n$ such that $a_x-a_{x+1}<0$, and thus there exist distinct $1\le p,q\le n$ such that $a_p-a_{p+1}>0$, $a_{p+2}-a_{p+3}<0$ and $a_q-a_{q+1}<0$, $a_{q+2}-a_{q+3}>0$.

Now we consider the value of $|a_i-a_{i+1}+a_{i+2}-a_{i+3}|$. If $a_i-a_{i+1}$ and $a_{i+2}-a_{i+3}$ have the same sign then $|a_i-a_{i+1}+a_{i+2}-a_{i+3}|=|a_i-a_{i+1}|+|a_{i+2}-a_{i+3}|$, and if $a_i-a_{i+1}$ and $a_{i+2}-a_{i+3}$ have different signs then $$|a_i-a_{i+1}+a_{i+2}-a_{i+3}|=||a_i-a_{i+1}|-|a_{i+2}-a_{i+3}||\le |a_i-a_{i+1}|+|a_{i+2}-a_{i+3}|-2.$$ So ${\sum }_{i=1}^n|a_i-a_{i+1}+a_{i+2}-a_{i+3}|$ is at most $$\begin{array}{rl}& \sum _{1\le i\le n,i\ne p,q}(|a_i-a_{i+1}|+|a_{i+2}-a_{i+3}|)\\ & +|a_p-a_{p+1}|+|a_{p+2}-a_{p+3}|+|a_q-a_{q+1}|+|a_{q+2}-a_{q+3}|-4\\ & =2\sum _{i=1}^n|a_i-a_{i+1}|-4.\end{array}$$ Let $k=\frac{n+1}{2}$. Then, $$2\sum _{i=1}^n|a_i-a_{i+1}|-4\le 2\sum _{i=1}^n(|k-a_i|+|k-a_{i+1}|)-4=4\sum _{i=1}^n|k-a_i|-4=n^2-5.$$ Therefore, $$\sum _{i=1}^n|a_i-a_{i+1}+a_{i+2}-a_{i+3}|\le n^2-5$$ and the equality is attained when $a_n=\frac{n+1}{2}$ and $a_{2i}=i$, $a_{2i-1}=\frac{n+1}{2}+i$ for $i=1,2,\dots ,\frac{n-1}{2}$. Thus the maximum of ${\sum }_{i=1}^n|a_i-a_{i+1}+a_{i+2}-a_{i+3}|$ is $n^2-5$. $\square$