A number Theory question on reaching 100,000 points

First we show that $n \le 48.$ For this, it suffices to exhibit a set of $48$ positive integers with the desired property.

So let $s_1 = 2, s_2 = 3, \ldots, s_{44} = 45; s_{45} = 64, s_{46} = 65, s_{47} = 66, s_{48} = 67.$ Then $\prod_{i=1}^{48} \left( 1-\frac1{s_i}\right) = \frac1{45} \cdot \frac{63}{67} = \frac{63}{3015} = \frac{42}{2010},$ as desired.

Now we want to give a lower bound for $n.$ The idea is that $\frac{42}{2010}$ is so small that we need a lot of terms in the product.

Suppose without loss of generality that the $s_i$ are in ascending order. Note that $s_i \ge i+1$ (since $s_1=1$ is impossible). Then $\prod_{i=1}^n \left( 1-\frac1{s_i} \right) \ge \prod_{i=1}^n \left( 1-\frac1{i+1} \right) = \frac12 \cdot \frac23 \cdots \frac{n}{n+1} = \frac1{n+1}.$

Since $\frac{42}{2010} \ge \frac1{n+1},$ this already shows that $n \ge 46.85...$ This bound is not quite good enough.

We can tighten this a bit in our case because one of the $s_i$ must be divisible by the prime $67$ since the denominator of the product is. The minimum of the product with respect to this requirement is clearly when $s_i = i+1$ for $1 \le i \le n-1$ and $s_n = 67.$ So $\prod_{i=1}^n \left( 1-\frac1{s_i} \right) \ge \frac1{n} \cdot \frac{66}{67}.$ So $\frac{42}{2010} \ge \frac{66}{67n},$ which gives $n \ge \frac{30 \cdot 66}{42} = 47.14...$

Putting the lower and upper bounds together shows that $n=\fbox{48}.$