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,
$$2 \left( \sum_{i=1}^{n} a_i^2 - \sum_{i=1}^{n} a_i a_{i+1} \right) = \sum_{i=1}^{n} (a_i - a_{i+1})^2 \\ \le (a_1 - a_3)^2 + \sum_{i=3}^{n} (a_i - a_{i+1})^2,$$
which transformed the question into the case of $2k$ numbers. We have
$$2 \left( \sum_{i=1}^{n} a_i^2 - \sum_{i=1}^{n} a_i a_{i+1} \right) \le (a_1 - a_3)^2 + \sum_{i=3}^{n} (a_i - a_{i+1})^2 \\ \le 2k (M - m)^2,$$
i.e.,
$$\left( \sum_{i=1}^{n} a_i^2 - \sum_{i=1}^{n} a_i a_{i+1} \right) \le k (M-m)^2 = \left[ \frac{n}{2} \right] (M-m)^2.$$