The least $n$

Clearly $s_i\neq 1$ for any $i$ . WLOG suppose $1<s_1<s_2<\cdots<s_n$ which implies $2\le s_1\le s_2-1\le s_3-2\le\cdots\le s_n-(n-1)$ hence for all $j$ we have $s_j\ge j+1$ . This gives $\dfrac{51}{2010}=\prod_{j=1}^n\left(1-\dfrac{1}{s_j}\right)\ge \prod_{j=1}^n\left(1-\dfrac{1}{j+1}\right)=\prod_{j=1}^n\dfrac{j}{j+1}=\dfrac{1}{n+1}\implies n\ge 39.$

It remains to show that $n=39$ works. Left to the reader as an exercise.

This is ISL 2010 N1 by the way.