Assume the contrary. The sum of the numbers is
$$\frac{2n^2m(n^2m-1)}{(m^2+n^2)(m^2-n^2)}$$
Since $m$ and $n$ are relatively prime, we have
$$\gcd(n^2m, m^2+n^2) = \gcd(n^2m, m^2-n^2) = 1.$$
Then we obtain $(m^2+n^2)(m^2-n^2) \mid 2(n^2m-1)$. Note that $m, n \ge 1$ and we can not have $m=n=1$, and hence $n^2m-1 > 0$. Therefore, $|(m^2+n^2)(m^2-n^2)| \le 2(n^2m-1) < 2n^2m$. As $m^2+n^2 > n^2$, we get $|m^2-n^2| < 2m$, and thus, $(m-1)^2-1 < n^2 < (m+1)^2-1$. So we have $n^2 = m^2$ or $n^2 = (m-1)^2$. It is clear that $n^2 = m^2$ is impossible. Hence, $n^2 = (m-1)^2$ which yields $n = m-1$. But in that case we obtain that one of the given numbers is not an integer, contradiction.