Harmony

The sum is $\begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2} \; = \; \sum_{n \ge m \ge 1} \frac{1}{m^2n^2} & = & \displaystyle \sum_{n > m \ge 1} \frac{1}{m^2n^2} + \sum_{n=1}^\infty \frac{1}{n^4} \; = \; \tfrac12\sum_{m \neq n} \frac{1}{m^2n^2} + \zeta(4) \\ & = & \displaystyle \frac12\left[\left(\sum_{n=1}^\infty \frac{1}{n^2}\right)^2 - \sum_{n=1}^\infty \frac{1}{n^4}\right] + \zeta(4) \\ & = & \tfrac12\big[\zeta(2)^2 + \zeta(4)\big] \; =\; \tfrac{7}{360}\pi^4 \end{array}$ making the answer $7 + 360 + 4 = \boxed{371}$ .

The sum can be expressed as

$\sum_{n=1}^{\infty}\sum_{m=1}^{n}\frac{1}{(mn)^2}$

Rearranging the terms, we get

$\sum_{n=1}^{\infty}\sum_{m=1}^{n}\frac{1}{(mn)^2}=\sum_{n=1}^{\infty}\frac{1}{n^2}+\sum_{n=2}^{\infty}\frac{1}{(2n)^2}+\sum_{n=3}^{\infty}\frac{1}{(3n)^2}...$

$=\sum_{k=1}^{\infty}\frac{1}{k^2}\sum_{n=k}^{\infty}\frac{1}{n^2}=\sum_{k=1}^{\infty}\frac{1}{k^2}\left(\zeta(2)-\sum_{n=1}^{k-1}\frac{1}{k^2}\right)$

$=\zeta(2)^2-\sum_{n=1}^{\infty}\frac{H^{(2)}_{n-1}}{n^2}$

Since $H^{(2)}_{n-1}=H^{(2)}_{n}-\frac{1}{n^2}$

$\sum_{n=1}^{\infty}\frac{H^{(2)}_{n-1}}{n^2}=\sum_{n=1}^{\infty}\frac{H^{(2)}_{n}}{n^2}-\frac{1}{n^4}=\sum_{n=1}^{\infty}\frac{H^{(2)}_{n}}{n^2}-\zeta(4)$

Putting it all together

$\sum_{n=1}^{\infty}\sum_{m=1}^{n}\frac{1}{(mn)^2}=\zeta(2)^2-\sum_{n=1}^{\infty}\frac{H^{(2)}_{n}}{n^2}+\zeta(4)$

$\sum_{n=1}^{\infty}\frac{H^{(2)}_{n}}{n^2}=\zeta(2)^2-\sum_{n=1}^{\infty}\frac{H^{(2)}_{n}}{n^2}+\zeta(4)$

$\sum_{n=1}^{\infty}\frac{H^{(2)}_{n}}{n^2}=\frac{1}{2}\left(\zeta(2)^2+\zeta(4)\right)$

Substituting $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta(4)=\frac{\pi^4}{90}$ , we get

$\sum_{n=1}^{\infty}\frac{H^{(2)}_{n}}{n^2}=\boxed{\frac{7\pi^4}{360}}$