https://artofproblemsolving.com/community/q3h2675724p23191612

Solution. Generalized wrote: We divide the set $\{1,2,\cdots,2n\}$ into two disjoint sets : $A=\{a_1,a_2,\cdots,a_n\}$ and $B=\{b_1,b_2,\cdots,b_n\}$ such that : $$a_1<a_2<\cdots<a_n\text{ and } b_1>b_2>\cdots>b_n.$$Show that : $$|a_1-b_1|+\cdots+|a_n-b_n|=n^2. $$ We use induction. The base case $n=1$ is trivial. Assume that it's true for $n=k \ge 1$. We show that it's true for $n=k+1$. WLOG $b_{1}=2k+2$. a. If $a_{1} = 1$, we're done since $$|a_1-b_1|+\cdots+|a_{k+1}-b_{k+1}|=(2k+1) + (k)^2 = (k+1)^2$$b. If $a_{1} \neq 1$, let $a_{1} = c > 1$. Note that $|a_1-b_1| = 2(k+1)-c$. Note that we can't apply the remaining $(2k+2) - 2 = 2k$ numbers using the induction hypothesis since they're not consecutive. We make them consecutive by shifting $1,2,\dots, c-1$ (which all are in $B$ by the minimality of $a_1$), a step forward. This would result in $$|a_1-b_1|+\cdots+|a_{k+1}-b_{k+1}| = 2(k+1)-c + (k)^2 + (c-1) = (k+1)^2$$