jmc

The fact that $x\equiv 0\mathrm{mod}7\Rightarrow 7\mid x$ is assumed as common knowledge in this answer. First, note that there are $8$ possible numbers that are equivalent to $1\mathrm{mod}7$, and there are $7$ possible numbers equivalent to each of $2$-$6\mathrm{mod}7$. Second, note that there can be no pairs of numbers $a$ and $b$ such that $a\equiv -b$ mod $7$, because then $a+b|7$. These pairs are $(0,0)$, $(1,6)$, $(2,5)$, and $(3,4)$. Because $(0,0)$ is a pair, there can always be $1$ number equivalent to $0\mathrm{mod}7$, and no more. To maximize the amount of numbers in S, we will use $1$ number equivalent to $0\mathrm{mod}7$, $8$ numbers equivalent to $1$, and $14$ numbers equivalent to $2$-$5$. This is obvious if you think for a moment. Therefore the answer is $1+8+14=\boxed{23}$ numbers.