Least $n$ .

Since $\prod_{j=1}^n \frac{s_j-1}{s_j} \; = \; \frac{42}{2010} \; = \; \frac{7}{5 \times 67}$ without loss of generality we deduce that $67$ must divide $s_n$ , and hence $s_n \ge 67$ . We can also assume that $2 \le s_1 < s_2 < ... < s_{n-1}$ so that $\frac{42}{2010} \; = \; \prod_{j=1}^n \big(1 - s_j^{-1}\big) \; \ge \; \frac12 \times \frac23 \times \frac34 \times \frac 45 \times ... \times \frac{n-1}{n} \times \frac{66}{67} \; = \; \frac{66}{67n}$ and hence $n \; \ge \; \frac{66 \times 2010}{42 \times 67} \; = \; 47.14$ so that $n \ge 48$ . But since $\frac{42}{2010} \; = \; \frac12 \times \frac{2}{3} \times ... \times \frac{32}{33} \times \frac{35}{36} \times \frac{36}{37} \times ... \times \frac{49}{50} \times \frac{66}{67}$ we see that $n = \boxed{48}$ is possible.