jmc

The given information tells us that \begin{array}{c@{\qquad}c} h(2)=j(2)=2, & h(4)=j(4)=6, \\ h(6)=j(6)=12, & h(8)=j(8)=12. \end{array}If the graphs of $y=h(2x)$ and $y=2j(x)$ intersect at $(a,b),$ then $$h(2a)=2j(a)=b.$$Checking the possibilities in the table above, we see that $h(8)=2j(4)=12.$ Thus, the graphs of $y=h(2x)$ and $y=2j(x)$ intersect at $(4,12),$ the sum of whose coordinates is $\boxed{16}.$