9. 82 Let $m$ and $n$ be two coprime natural numbers and $n < m$. Try to compare the following two numbers:
$$\sum_{k=1}^{n}\left[k \cdot \frac{m}{n}\right] \text { and } \sum_{k=1}^{m}\left[k \cdot \frac{n}{m}\right]$$

in size.
[Solution] Since
$$\frac{k m}{n}=\left[\frac{k m}{n}\right]+\frac{a_{k}}{n}, k=1,2, \cdots, n$$

where $a_{k}$ is an integer and $0 \leqslant a_{k} \leqslant n-1$. Also, since $m, n$ are coprime, $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ is a permutation of $\{0,1,2, \cdots, n-1\}$. Therefore,
$$\begin{aligned}
\sum_{k=1}^{n}\left[k \cdot \frac{m}{n}\right] & =\sum_{k=1}^{n} \frac{k m}{n}-\sum_{k=1}^{n} \frac{a_{k}}{n} \\
& =\frac{1}{2} m(n+1)-\frac{1}{2}(n-1) \\
& =m+\frac{1}{2}(m-1)(n-1)
\end{aligned}$$

Similarly, we get $\quad \sum_{k=1}^{m}\left[k \cdot \frac{n}{m}\right]=n+\frac{1}{2}(m-1)(n-1)$,
so $\quad \sum_{k=1}^{n}\left[k \cdot \frac{m}{n}\right]>\sum_{k=1}^{m}\left[k \cdot \frac{n}{m}\right]$.