Who's up to the challenge? 28

Using the symmetry in $x$ and $y$ in the integrand, $\begin{array}{rcl} I & = & \displaystyle \int_0^1 \int_0^1 \left\{\frac{k}{x-y}\right\} \left\{\frac{1}{x}\right\}\left\{\frac{1}{y}\right\}\,dx\,dy \\ & = & \displaystyle \frac12\int_0^1 \int_0^1\left( \left\{\frac{k}{x-y}\right\} + \left\{ \frac{k}{y-x}\right\} \right) \left\{\frac{1}{x}\right\}\left\{\frac{1}{y}\right\}\,dx\,dy \\ & = & \displaystyle\frac12 \int_0^1 \int_0^1 \left\{\frac{1}{x}\right\} \left\{\frac{1}{y}\right\}\,dx\,dy \\ & = & \displaystyle \frac12\left(\int_0^1 \big\{ x^{-1}\big\}\,dx\right)^2 \end{array}$ using the fact that $\{u\} + \{-u\} = 1$ . Then $\begin{array}{rcl} \displaystyle \int_0^1 \big\{ x^{-1} \big\}\,dx & = & \displaystyle \int_1^\infty \big\{ y \big\} y^{-2}\,dy \\ & = & \displaystyle \sum_{n=1}^\infty \int_0^1 \frac{y}{(n+y)^2} \,dy \; = \; \sum_{n=1}^\infty \Big[ \ln(n+y) + \frac{n}{n+y} \Big]_0^1 \\ & = & \displaystyle \lim_{N \to \infty}\sum_{n=1}^N \left( \ln(n+1) - \ln n - \frac{1}{n+1} \right) \\ & = & \displaystyle \lim_{N \to \infty} \left( \ln(N+1) - \sum_{n=1}^N \frac{1}{n+1} \right) \; =\; 1 - \gamma \;. \end{array}$ Thus $I \,=\, \tfrac12(1 - \gamma)^2$ , making the answer $1 + 2 + 1 + 1 + 1 + 1 + 2 = \boxed{8}$ .