jmc

Multiplying the second equation by $i$ and adding the first equation, we get $$a+bi+\frac{17a+6b+6ai-17bi}{a^2+b^2}=6.$$We can write $$\begin{array}{rl}17a+6b+6ai-17bi& =(17+6i)a+(6-17i)b\\ & =(17+6i)a-(17+6i)bi\\ & =(17+6i)(a-bi).\end{array}$$Also, $a^2+b^2=(a+bi)(a-bi),$ so $$a+bi+\frac{(17+6i)(a-bi)}{(a+bi)(a-bi)}=6.$$This simplifies to $$a+bi+\frac{17+6i}{a+bi}=6.$$Let $z=a+bi,$ so $$z+\frac{17+6i}{z}=6.$$This becomes $z^2-6z+(17+6i)=0.$ By the quadratic formula, $$z=\frac{6\pm \sqrt{36-4(17+6i)}}{2}=\frac{6\pm \sqrt{-32-24i}}{2}=3\pm \sqrt{-8-6i}.$$We want to find the square roots of $-8-6i,$ so let $$-8-6i=(u+vi{)}^2=u^2+2uvi+v^2{i}^2=u^2+2uvi-v^2.$$Equating the real and imaginary parts, we get $u^2-v^2=-8$ and $2uv=-6,$ so $uv=-3.$ Then $v=-\frac{3}{u}.$ Substituting, we get $$u^2-\frac{9}{u^2}=-8.$$Then $u^4+8u^2-9=0,$ which factors as $(u^2-1)(u^2+9)=0.$ Hence, $u=1$ or $u=-1.$ If $u=1,$ then $v=-3.$ If $u=-1,$ then $v=3.$ Thus, the square roots of $-8-6i$ are $1-3i$ and $-1+3i.$

For the square root $1-3i,$ $$z=3+1-3i=4-3i.$$This gives the solution $(a,b)=(4,-3).$

For the square root $-1+3i,$ $$z=3-1+3i=2+3i.$$This gives the solution $(a,b)=(2,3).$

The final answer is then $4+(-3)+2+3=\boxed{6}.$