Solution:

Let $A_{o}, B_{o}, C_{o}$ be sets satisfying the said conditions. Note that $1 \in A_{o}$ implies that $1 \in B_{o}$ and $1 \notin S_{2006}-A_{o}$ so that 1 may or may not be in $C_{o}$. Also,
$1 \notin A_{o}$ implies that $1 \in S_{2006}-A_{o} \subset C_{o}$ while 1 may or may not be in $B_{o}$. Thus there are four possibilities for the distribution of 1, and since the same argument holds independently for $2,3, \ldots, 2006$, the answer is $4^{2006}$ or $2^{4012}$.