jmc

Note that it is impossible for any of $h,t,u$ to be $1$, since then each picket will have been painted one time, and then some will be painted more than once. $h$ cannot be $2$, or that will result in painting the third picket twice. If $h=3$, then $t$ may not equal anything not divisible by $3$, and the same for $u$. Now for fourth and fifth pickets to be painted, $t$ and $u$ must be $3$ as well. This configuration works, so $333$ is paintable. If $h$ is $4$, then $t$ must be even. The same for $u$, except that it can't be $2\mathrm{mod}4$. Thus $u$ is $0\mathrm{mod}4$ and $t$ is $2\mathrm{mod}4$. Since this is all $\mathrm{mod}4$, $t$ must be $2$ and $u$ must be $4$, in order for $5,6$ to be paint-able. Thus $424$ is paintable. $h$ cannot be greater than $5$, since if that were the case then the answer would be greater than $999$, which would be impossible for the AIME. Thus the sum of all paintable numbers is $\boxed{757}$.