jmc

If the graphs intersect at $(a,b)$, then we have $$h(a)=h(a-3)(=b).$$Thus, $(a,b)$ and $(a-3,b)$ are both on the original graph of $y=h(x)$. Looking for two points on the original graph which are separated by $3$ units horizontally, we find $(-2,3)$ and $(1,3)$. Thus $a-3=-2,$ $a=1,$ and $b=3;$ the graphs of $y=h(x)$ and $y=h(x-3)$ intersect at $(1,3),$ the sum of whose coordinates is $\boxed{4}$.