jmc

Let us first analyze the fraction $\frac{17k}{66}$. We can rewrite this fraction as $\frac{17k}{2\cdot 3\cdot 11}$. Since the denominator can only contain powers of 2 and 5, we have that $k$ must be a multiple of 33. We now continue to analyze the fraction $\frac{13k}{105}$. We rewrite this fraction as $\frac{13k}{3\cdot 5\cdot 7}$, and therefore deduce using similar logic that $k$ must be a multiple of 21. From here, we proceed to find the least common multiple of 21 and 33. Since $21=3\cdot 7$ and $33=3\cdot 11$, we conclude that the least common multiple of 21 and 33 is $3\cdot 7\cdot 11=231$.

We now know that $S$ contains exactly the multiples of 231. The smallest multiple of 231 that is greater than 2010 is $231\cdot 9=\boxed{2079}$.