Cute looking nasty summation!

Note that $\begin{array}{rcl} \displaystyle \sum_{k=1}^\infty \frac{H_k^3}{k^4} & = & \displaystyle 1 + \sum_{k=1}^\infty \frac{H_{k+1}^3}{(k+1)^4} \; = \; 1 + \sum_{k=1}^\infty \frac{1}{(k+1)^4}\big[H_k + \tfrac{1}{k+1}\big]^3 \\ & = & \displaystyle 1 + \sum_{k=1}^\infty \left(\frac{H_k^3}{(k+1)^4} + 3\frac{H_k^2}{(k+1)^5} + 3\frac{H_k}{(k+1)^6} + \frac{1}{(k+1)^7}\right) \\ & = & s_h(3,4) + 3s_h(2,5) + 3s_h(1,6) + \zeta(7) \end{array}$ where we are using the Euler sums $s_h(m,n) \; = \; \sum_{k=1}^\infty \frac{H_k^m}{(k+1)^n} \;,$ it can be shown by standard results concerning Euler sums (formulae for $s_h(1,n)$ , $s_h(2,2n-1)$ and $s_h(3,4)$ are all known in terms of Zeta functions) that $\sum_{k=1}^\infty \frac{H_k^3}{k^4} \; =\; \tfrac{231}{16}\zeta(7) - \tfrac{51}{4}\zeta(3)\zeta(4) + 2\zeta(2)\zeta(5) \;,$ making the answer $231 + 16 + 7 + 51 + 4 +3 + 4 + 2 + 2 + 5 \,=\, \boxed{325}$ .