48th Austrian Mathematical Olympiad

Answer: $n=2^{2016}$.

Proof: There are $2^{2016}$ subsets of $S$ which do not contain $2017$. The union of any two such subsets does not contain $2017$ and is thus a proper subset of $S$. Thus $n\ge 2^{2016}$.

To show the other direction, we group the subsets of $S$ into $2^{2016}$ pairs in such a way that every subset forms a pair with its complement. If $n>2^{2016}$ then the $n$ subsets would contain such a pair. Its union would be $S$, contradiction.

Thus $n=2^{2016}$.