So, suppose we have a counterexample. Every positive integer $n$ can be expressed as the product of an odd number and a power of 2, say $2^{k}\times \ell$ for odd $\ell$, nonnegative $k$. Classify integers in $S$ by the odd number $\ell$ obtained above, obtaining $1006$ classes. We cannot have any two numbers are in the same class with the same odd $\ell$: certainly, $2^{k_1}\times \ell \mid 2(2^{k_2}\times \ell)$ if $k_1 < k_2$. At the same time, for an odd integer $505 \leq m \leq 669$, there is a class $\{m, 2m\}$ and a class $\{3m\}$. We have $m \mid 2(3m)$ and $2m \mid 2(3m)$, so at most one of these classes can contain a number. Then at least $\frac{669-505}{2}+1 = 83$ classes must be empty, so $923$ classes are left, certainly not enough for all $1000$ elements.