Taiwan IMO Selection Camp

Answer: $n = 39$.

Without loss of generality, assume $s_1 < s_2 < \cdots < s_n$. Clearly, $s_1 > 1$, otherwise $1 - \frac{1}{s_1} = 0$. Therefore, $2 \le s_1 \le s_2 - 1 \le \cdots \le s_n - (n-1)$. Hence $s_i \ge i + 1, \forall i = 1, \cdots, n$. From the given condition we obtain
$$\begin{aligned} \frac{17}{670} &= (1 - \frac{1}{s_1})(1 - \frac{1}{s_2})\cdots(1 - \frac{1}{s_n}) \\ &\ge (1 - \frac{1}{2})(1 - \frac{1}{3})\cdots(1 - \frac{1}{n+1}) \\ &= (1 - \frac{1}{2})(1 - \frac{1}{3})\cdots(1 - \frac{1}{n+1}) = \frac{1}{n+1} \\ n + 1 &\ge \frac{670}{17} > 39 \\ n &\ge 39. \end{aligned}$$

Next we prove that $n = 39$ satisfies the requirement.

Consider the 39 distinct positive integers: $2, 3, \dots, 33, 35, 36, 40, 67$. We obtain
$$\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{32}{33} \cdot \frac{34}{35} \cdot \frac{39}{40} \cdot \frac{66}{67} = \frac{17}{67}.$$
This completes the proof.