Solution:
Assume without loss of generality that the minimum element of $S$ is $0$. By the previous problem, $S$ tiles the set $\{1,2, \ldots, k\}$ for some positive integer $k$. Then let $P(x)$ be the polynomial $\sum_{i=0}^{n} x^{s_{i}}$. To say that the set $\{1,2, \ldots, k\}$, or equivalently the set $\{0,1, \ldots, k-1\}$, is tiled by $S$ is to say that there is some polynomial $Q(x)$ with coefficients $0$ or $1$ such that $P(x) Q(x) = 1 + x + \cdots + x^{k-1} = \frac{x^{k}-1}{x-1}$. It follows that all the roots of $P(x)$ are roots of unity, but $P(1) \neq 0$. By question 11 above, this implies that $P(x)$ is symmetric. Therefore, $s_{0} + s_{n} = s_{1} + s_{n-1} = \cdots = s_{n} + s_{0}$, so $S$ is symmetric.