Riemann everywhere

Starting with, $\displaystyle S=\sum_{n\ge 1} \frac{\psi^{(1)}(n)}{n^2} = \sum_{n\ge 1} \frac{\zeta(2)-H_{n-1}^{(2)}}{n^2} = \zeta^2 (2) - \sum_{n\ge 1}\frac{H_{n}^{(2)}-\frac{1}{n^2}}{n^2} = \zeta^2 (2) + \zeta(4) - \sum_{n\ge 1} \frac{H_{n}^{(2)}}{n^2}$

Using the generalised harmonic summation , $\displaystyle \sum_{n=1}^{\infty} \frac{H_{n}^{(m)}}{n^m} = \frac{1}{2}( \zeta^2(m) +\zeta(2m))$ for $m\ge 2$

Using above we have, $\displaystyle S= \frac{1}{2} (\zeta^2(2) + \zeta(4)) = \frac{7\pi^4}{360}=\frac{7}{4}\zeta(4)$

Hence the answer : $\boxed{7+4+4=15}$