jmc

If $a=0,$ then $\frac{10}{b}=5,$ so $b=2,$ which does not satisfy the second equation. If $b=0,$ then $\frac{10}{a}=4,$ so $a=\frac{5}{2},$ which does not satisfy the first equation. So, we can assume that both $a$ and $b$ are nonzero.

Then $$\frac{5-a}{b}=\frac{4-b}{a}=\frac{10}{a^2+b^2}.$$Hence, $$\frac{5b-ab}{b^2}=\frac{4a-ab}{a^2}=\frac{10}{a^2+b^2},$$so $$\frac{4a+5b-2ab}{a^2+b^2}=\frac{10}{a^2+b^2},$$so $4a+5b-2ab=10.$ Then $2ab-4a-5b+10=0,$ which factors as $(2a-5)(b-2)=0.$ Hence, $a=\frac{5}{2}$ or $b=2.$

If $a=\frac{5}{2},$ then $$\frac{5/2}{b}=\frac{10}{\frac{25}{4}+b^2}.$$This simplifies to $4b^2-16b+25=0.$ By the quadratic formula, $$b=2\pm \frac{3i}{2}.$$If $b=2,$ then $$\frac{2}{a}=\frac{10}{a^2+4}.$$This simplifies to $a^2-5a+4=0,$ which factors as $(a-1)(a-4)=0,$ so $a=1$ or $a=4.$

Hence, the solutions are $(1,2),$ $(4,2),$ $\left(\frac{5}{2},2+\frac{3i}{2}\right),$ $\left(\frac{5}{2},2-\frac{3i}{2}\right),$ and the final answer is $$1+2+4+2+\frac{5}{2}+2+\frac{3i}{2}+\frac{5}{2}+2-\frac{3i}{2}=\boxed{18}.$$