China Mathematical Olympiad

If $n=2k$ ($k$ is a positive integer), then $$2\left(\sum _{i=1}^n{a}_i^2-\sum _{i=1}^n{a}_i{a}_{i+1}\right)=\sum _{i=1}^n(a_i-a_{i+1}{)}^2\le n(M-m{)}^2,$$ therefore, $$\sum _{i=1}^n{a}_i^2-\sum _{i=1}^n{a}_i{a}_{i+1}\le \frac{n}{2}(M-m{)}^2=\left[\frac{n}{2}\right](M-m{)}^2.$$ If $n=2k+1$ ($k$ is a positive integer), then for $2k+1$ numbers arranged in a cyclic way, one can always find three consecutive increasing or decreasing terms (as ${\prod }_{i=1}^{2k+1}(a_i-a_{i-1})(a_{i+1}-a_i)$), so it is not possible that for every $i$, $a_i-a_{i-1}$ and $a_{i+1}-a_i$ having opposite signs). Without loss of generality, we assume that $a_1,a_2,a_3$ are monotonic, then $$(a_1-a_2{)}^2+(a_2-a_3{)}^2\le (a_1-a_3{)}^2.$$ Hence, $$\begin{gathered} 2\left(\sum _{i=1}^n{a}_i^2-\sum _{i=1}^n{a}_i{a}_{i+1}\right)=\sum _{i=1}^n(a_i-a_{i+1}{)}^2 \\ \le (a_1-a_3{)}^2+\sum _{i=3}^n(a_i-a_{i+1}{)}^2, \end{gathered}$$ which transformed the question into the case of $2k$ numbers. We have $$\begin{gathered} 2\left(\sum _{i=1}^n{a}_i^2-\sum _{i=1}^n{a}_i{a}_{i+1}\right)\le (a_1-a_3{)}^2+\sum _{i=3}^n(a_i-a_{i+1}{)}^2 \\ \le 2k(M-m{)}^2, \end{gathered}$$ i.e., $$\left(\sum _{i=1}^n{a}_i^2-\sum _{i=1}^n{a}_i{a}_{i+1}\right)\le k(M-m{)}^2=\left[\frac{n}{2}\right](M-m{)}^2.$$