jmc

We begin by finding the equation of the line $\ell$ containing $(1,7)$ and $(13,16)$. The slope of $\ell$ is $\frac{16-7}{13-1}=\frac{9}{12}=\frac{3}{4}$, so the line has the point-slope form $y-7=\frac{3}{4}(x-1)$. Substituting the value $x=5$, we obtain that $y=7+\frac{3}{4}(5-1)=10$. It follows that the point $(5,10)$ lies on the line containing $(1,7)$ and $(13,16)$ (for $k=10$, we obtain a degenerate triangle). To minimize the area of the triangle, it follows that $k$ must either be equal to $9$ or $11$.

Indeed, we claim that both such triangles have the same area. Dropping the perpendiculars from $(5,9)$ and $(5,11)$ to $\ell$, we see that the perpendiculars, $\ell$, and the line segment connecting $(5,9)$ to $(5,11)$ form two right triangles. By vertical angles, they are similar, and since they both have a hypotenuse of length $1$, they must be congruent. Then, the height of both triangles must be the same, so both $k=9$ and $k=11$ yield triangles with minimal area. The answer is $9+11=\boxed{20}$.