Any three positive integers that multiply to make $2009$ would create viable cuboids.

The prime factors of $2009$ are $7\times 7\times 41$, so the options are:

$1 \times1 \times 2009$

$1\times 7\times 287$

$1\times 41\times 49$

$7\times 7 \times 41$

The first three cuboids all have two faces which each require $2009$ stickers ($1\times2009$, $7\times287$ and $41\times49$ respectively) so Ruth cannot cover them.

The last cuboid has surface area: $2\times( 7\times7+7\times41 + 41\times 7) = 1246$

This leaves $2009-1246=763$ stickers left over.

*This problem is taken from the UKMT Mathematical Challenges.*