10. Given that $m$ and $n$ are positive integers. If $1 \leqslant m \leqslant n$ $\leqslant 30$, and $mn$ is divisible by 21, then the number of pairs $(m, n)$ that satisfy the condition is $\qquad$ .
10.57.

Given that positive integers $m, n$ satisfy $mn$ is divisible by 21, and $1 \leqslant m \leqslant n \leqslant 30$, therefore,
(1) If $m=21$, then $n=21,22, \cdots, 30$. Hence, there are 10 pairs $(m, n)$ that satisfy the condition.
(2) If $m \neq 21$,
(i) When $n=21$, $m=1,2, \cdots, 20$. There are 20 pairs $(m, n)$ that satisfy the condition.
(ii) When $n \neq 21$, since $21=3 \times 7$, therefore,
1) If $m=3a, n=7b\left(a, b \in \mathbf{N}_{+}\right.$, and $a \neq 7, b \neq 3$), we get $1 \leqslant 3a \leqslant 7b \leqslant 30$.
When $b=1$, $a=1,2$;
When $b=2$, $a=1,2,3,4$;
When $b=4$, $a=1,2,3,4,5,6,8,9$.
Hence, there are
$2+4+8=14$ (pairs) that satisfy the condition.
2) If $m=7a, n=3b\left(a, b \in \mathbf{N}_{+}\right.$, and $a \neq 3, b \neq 7$), we get $1 \leqslant 7a \leqslant 3b \leqslant 30$.
When $b=3,4$, $a$ is 1;
When $b=5,6,8,9$, $a$ is 1,2;
When $b=10$, $a$ is 1,2,4.
Hence, there are
$$
2 \times 1+4 \times 2+3=13 \text{ (pairs). }
$$

In summary, there are
$$
10+20+14+13=57 \text{ (pairs) }
$$
that satisfy the condition.