Solution:

Suppose $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ are distinct positive integers that satisfy the given equation. Without loss of generality, we assume that $x_{1}<x_{2}<x_{3}<\cdots<x_{n}$. Then
$$2 \leq x_{1} \leq x_{2}-1 \leq x_{3}-2 \leq \cdots \leq x_{n}-(n-1)$$
and so $x_{i} \geq i+1$ for $1 \leq i \leq n$.
$$\begin{aligned} \frac{15}{2013} & =\left(1-\frac{1}{x_{1}}\right)\left(1-\frac{1}{x_{2}}\right)\left(1-\frac{1}{x_{3}}\right) \cdots\left(1-\frac{1}{x_{n}}\right) \\ & \geq\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \cdots\left(1-\frac{1}{n+1}\right) \\ & =\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n}{n+1} \\ & =\frac{1}{n+1} \end{aligned}$$
The preceding computation gives $n \geq 134$.

It remains to show that $n=134$ can be attained. Set $x_{i}=i+1$ for $1 \leq i \leq 133$, and $x_{134}=671$. Then
$$\left(1-\frac{1}{x_{1}}\right)\left(1-\frac{1}{x_{2}}\right)\left(1-\frac{1}{x_{3}}\right) \cdots\left(1-\frac{1}{x_{n}}\right)=\frac{1}{134} \cdot \frac{670}{671}=\frac{5}{671}=\frac{15}{2013}$$
Therefore, the required minimum value of $n$ is $134$.