$S$ is polygonal if and only if $2\max\{s_1, s_2, \ldots , s_k\} < s_1+s_2+\cdots +s_k$.

Consider the set $S$ of size $n$ defined as $S=\{2,4, \ldots ,2^{n-4}, 2^{n-3}, 2^{n-3}, 2^{n-3}, 2^{n-2}-1\}$. When this set is divided into two pieces either one set has three $2^{n-3}$'s, or one set has two $2^{n-3}$'s. In addition, that set can either have $2^{n-2}-1$ or not have it. No matter which of the 4 cases, it is easy to see that set is polygonal by the above hidden observation. The other set is not polygonal because it only contains distinct powers of two or near powers of two, and so by the geometric series summation formula the sum of all except the largest element is smaller than the largest element always, so by the observation again it is not polygonal. Thus, this set is multipolygonal. Now if we remove $2^{n-2}-1$ by almost identical logic the remaining set is also multipolygonal so this set works.