Japan competition problems, 2007

Now put $p = n - m$, $q = n$. Finding all positive integer pairs $(m, n)$ is equivalent to finding all integer pairs $(p, q)$. And then condition (1) and (3) is equivalent to the conditions that $1 \leq q \leq 20$, $\frac{1}{4} < \frac{p}{q} < \frac{2}{7}$. And condition (2) is equivalent to the condition that $p$ and $q$ are relatively prime. And $\frac{1}{4} < \frac{p}{q} < \frac{2}{7}$ is equivalent to $\frac{7}{2}p < q < 4p$. So,
* If $p \leq 0$, $q$ doesn't exist because of $\frac{p}{q} \leq 0$
* If $p = 1$, then $\frac{7}{2} < q < 4$, so $q$ doesn't exist.
* If $p = 2$, then $7 < q < 8$, so $q$ doesn't exist.
* If $p = 3$, then $\frac{21}{2} < q < 12$, so $q = 11$.
* If $p = 4$, then $14 < q < 16$, so $q = 15$.
* If $p = 5$, then $\frac{35}{2} < q < 20$, so $q = 18, 19$.
* If $p \geq 6$, $q$ doesn't exist because $q > \frac{7}{2}p \geq 21$.
Then all integer pairs $(p, q)$ are $(3, 11)$, $(4, 15)$, $(5, 18)$, $(5, 19)$. So, all positive integer pairs $(m, n)$ are $(13, 18)$, $(8, 11)$, $(11, 15)$, $(14, 19)$.