A bit of a madness (for juniors) with combinatorial thinking. Clearly $\{1\}$ is $0$-divided and $\{1,2,3\}$ is $1$-divided. For $n\geq 2$ work with $\{3n,3n+1,\ldots,4n-1,4n+1,4n+2,\ldots,5n,8n(n-2)\}$, the sum of all numbers being $16n^2 - 16n$. By placing $8n(n-2)$ in one part, we want all other elements in this part to have sum $8n$. As the smallest element is $3n$, there can be at most two other elements and now it is clear that the only option is $\{4n- \ell, 4n + \ell\}$, $\ell = 1,2,\ldots,n$.