Who's up to the challenge? 34

We have (putting $N = k+n$ ) $\begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{\zeta(k+n)-1}{k+n} & = & \displaystyle \sum_{N=2}^\infty \sum_{n=1}^{N-1} \frac{\zeta(N)-1}{N} \\ & = & \displaystyle \sum_{N=2}^\infty \frac{(N-1)(\zeta(N)-1)}{N} \; = \; \sum_{N=2}^\infty (\zeta(N)-1) - \sum_{N=2}^\infty \frac{\zeta(N)-1}{N} \\ & = & 1 - (1 - \gamma) \; = \; \gamma \end{array}$ making the answer $1 + 0 =\boxed{1}$ . Since all the terms in the series are positive, there is no problem rearranging the series in this manner.