[Number Theory]
Problem: Find the least positive integer $n$ for which there exists a set $\left\{s_{1}, s_{2}, \ldots, s_{n}\right\}$ consisting of $n$ distinct positive integers such that

$$
\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right)=\frac{51}{2010}
$$

Solution: Suppose that for some $n$ there exist the desired numbers; we may assume that $s_{1}<s_{2}<\cdots<s_{n}$. Surely $s_{1}>1$ since otherwise $1-\frac{1}{s_{1}}=0$. So we have $2 \leq s_{1} \leq s_{2}-1 \leq \cdots \leq s_{n}-(n-1)$, hence $s_{i} \geq i+1$ for each $i=1, \ldots, n$. Therefore

$$
\begin{aligned}
\frac{51}{2010} & =\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right) \\
& \geq\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \ldots\left(1-\frac{1}{n+1}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{n}{n+1}=\frac{1}{n+1}
\end{aligned}
$$

which implies

$$
n+1 \geq \frac{2010}{51}=\frac{670}{17}>39
$$

so $n \geq 39$.

Now we are left to show that $n=39$ fits. Consider the set $\{2,3, \ldots, 33,35,36, \ldots, 40,67\}$ which contains exactly 39 numbers. We have

$$
\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{32}{33} \cdot \frac{34}{35} \cdots \frac{39}{40} \cdot \frac{66}{67}=\frac{1}{33} \cdot \frac{34}{40} \cdot \frac{66}{67}=\frac{17}{670}=\frac{51}{2010}
\tag{1}
$$

hence for $n=39$ there exists a desired example.

Final answer: 39