Prove that since $m+n$ is an odd prime, $n-m+1$ is even. Therefore, $H(m, n)$ has an even number of terms.
$$\begin{array}{l} \text { Let } \frac{1}{m n}+\frac{1}{(m+1)(n-1)}+\cdots \\ \frac{m+n-1}{2} \cdot \frac{1}{2}+\frac{n+1}{2}=\frac{s}{r}, \\ (r, s)=1, r, s \in N \text {. } \\ \end{array}$$

It is easy to see that the prime factors of $r$ are all less than or equal to $n$, so,
$$\begin{array}{l} (m+n, r)=1 . \\ \because H(m, n)=\frac{q}{p}, \\ \therefore \frac{q}{p}=\frac{(m+n) s}{r}, \end{array}$$

which means
$$\begin{array}{l} (m+n) s p=r q . \\ \because(m+n, r)=1, \\ \therefore(m+n) \mid q . \end{array}$$

The 21st IMO has a problem as follows:
Let $p, q$ be natural numbers, $\frac{q}{p}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$ $-\frac{1}{1318}+\frac{1}{1319}$. Prove: $1979|q$.
It is easy to see that
$$\begin{aligned} \frac{q}{p} & =H(1,1319)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{1318}\right) \\ & =H(1,1319)-H(1,659) \\ & =H(660,1319) . \end{aligned}$$

Since $660+1319=1979$ is a prime number, by Example 2, we know $1979 \mid q$.