Too many things to add!

Since $\sum_{n=1}^\infty \frac{1}{n^2 + kn} \; =\; \frac{1}{k}\sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+k}\right) \; = \; \frac{H_k}{k}$ we need to evaluate $4\sum_{k=1}^\infty \frac{H_k^2}{k^2} \;.$ Now $\begin{array}{rcl} \displaystyle \sum_{k=1}^\infty \frac{H_k^2}{k^2} & = & \displaystyle \sum_{k=1}^\infty \frac{1}{k^2}\big(H_{k-1} + k^{-1}\big)^2 \\ & = & \displaystyle \sum_{k=1}^\infty \frac{H_k^2}{(k+1)^2} + 2\sum_{k=1}^\infty \frac{H_k}{(k+1)^3} + \sum_{k=1}^\infty \frac{1}{k^4} \\ & = & \displaystyle \sum_{a > b,c} \frac{1}{a^2bc} + 2\sum_{a>b}\frac{1}{a^3b} + \zeta(4) \\ & = & \displaystyle 2\sum_{a>b>c} \frac{1}{a^2bc} + \sum_{a>b}\frac{1}{a^2b^2} + 2\sum_{a>b} \frac{1}{a^3b} + \zeta(4) \\ & = & 2\zeta(2,1,1) + \zeta(2,2) + 2\zeta(3,1) + \zeta(4) \; = \; \tfrac{17}{360}\pi^4\;, \end{array}$ and so the sum is equal to $\tfrac{17}{90}\pi^4$ , making the answer $17 + 90 + 4 = \boxed{111}$ .