 Solution

Starting from $n$ singleton blocks, a partition of a set into $n-1$ nonempty blocks has exactly one block of size two and all other blocks of size one: the sum of the block sizes exceeds the number of blocks by exactly one. Choosing that pair determines the partition uniquely. HenceThis fixed value of $k$ works for every $n\ge 2$. It is forced already by $n=2$: $S(2,1)=1$, whereas for positive $k$ the only $\left(\genfrac{}{}{0ex}{}{2}{k}\right)$ equal to one is $\left(\genfrac{}{}{0ex}{}{2}{2}\right)$.