Chinese Mathematical Olympiad

Without loss of generality, we may assume that $a_1<a_2<\cdots <a_n$, and note also that $$|a_k|+|a_{n-k+1}|\ge |a_{n-k+1}-a_k|\ge n+1-2k$$ for $1\le k\le n$. So $$\begin{array}{rl}\sum _{k=1}^n|a_k{|}^3& =\frac{1}{2}\sum _{k=1}^n(|a_k{|}^3+|a_{n+1-k}{|}^3)\\ & =\frac{1}{2}\sum _{k=1}^n(|a_k{|}^3+|a_{n+1-k}|)(\frac{3}{4}(|a_k|-|a_{n+1-k}|{)}^2\\ & +\frac{1}{4}(|a_k{|}^3+|a_{n+1-k}|{)}^2)\\ & \ge \frac{1}{8}\sum _{k=1}^n(|a_k{|}^3+|a_{n+1-k}|{)}^3\\ & \ge \frac{1}{8}\sum _{k=1}^n|n+1-2k{|}^3.\end{array}$$ When $n$ is odd, $$\sum _{k=1}^n|n+1-2k{|}^3=2\times 2^3\times \sum _{i=1}^{\frac{n-1}{2}}{i}^3=\frac{1}{4}(n^2-1{)}^2.$$ When $n$ is even, $$\begin{array}{rl}\sum _{k=1}^n|n+1-2k{|}^3& =2\sum _{i=1}^{\frac{n}{2}}(2i-1{)}^3\\ & =2\left(\sum _{j=1}^n{j}^3-\sum _{i=1}^{\frac{n}{2}}(2i{)}^3\right)\\ & =\frac{1}{4}{n}^2(n^2-2).\end{array}$$ So $$\sum _{k=1}^n|a_k{|}^3\ge \frac{1}{32}(n^2-1{)}^2$$ for odd $n$, and $$\sum _{k=1}^n|a_k{|}^3\ge \frac{1}{32}{n}^2(n^2-2)$$ for even $n$. The equality holds at $a_i=i-\frac{n+1}{2}$, $i=1,2,\dots ,n$.