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Let $(x,y)$ be one such point. Since $(x,y)$ is $5$ units from the line $y=13$, it must be 5 units above the line or 5 units below it. This means that the $y$-coordinate is 8 or 18. By the distance formula, since $(x,y)$ is 13 units from $(7,13)$, we have $$\sqrt{(x-7{)}^2+(y-13{)}^2}=13.$$Since $y$ is either 5 more or 5 less than 13, we know that $(y-13{)}^2=25$. So we substitute:

$$\begin{array}{rl}\sqrt{(x-7{)}^2+25}& =13\\ \Rightarrow (x-7{)}^2+25& =13^2\\ \Rightarrow (x-7{)}^2& =144\\ \Rightarrow x-7& =\pm 12.\end{array}$$So we have $x-7=12$ or $x-7=-12$, giving $x=19$ or $x=-5$.

Putting it all together, we have $y=8\text{ or }18$ and $x=-5\text{ or }19$, so our four possible points are $(-5,8),$ $(-5,18),$ $(19,8),$ and $(19,18).$ The sum of all of these coordinates is $\boxed{80}$.