It is clear that $x \in \mathbb{Q}$ and $x > 0$.

On the other hand, if $x$ has such a representation, then from the inequalities
$$\begin{aligned} & \frac{1}{a_{1} a_{2} \ldots a_{n}} \leq \frac{1}{a_{2} a_{3} \ldots a_{n}} \\ & \frac{a_{1}}{a_{2} a_{3} \ldots a_{n}} \leq \frac{a_{2} - 1}{a_{2} a_{3} \ldots a_{n}} = \frac{1}{a_{3} a_{4} \ldots a_{n}} - \frac{1}{a_{2} a_{3} \ldots a_{n}} \\ & \frac{a_{n-2}}{a_{n-1} a_{n}} \leq \frac{a_{n-1} - 1}{a_{n-1} a_{n}} = \frac{1}{a_{n}} - \frac{1}{a_{n-1} a_{n}} \\ & \frac{a_{n-1}}{a_{n}} \leq \frac{a_{n} - 1}{a_{n}} = 1 - \frac{1}{a_{n}}, \end{aligned}$$
we get $x \leq 1 - \frac{1}{a_{n}}$, hence $x \in (0, 1]$. We note that $x = 1$ can be represented in this form by considering $n = 1$ and $a_{1} = 1$.

If $x = \frac{p}{q} \in (0, 1)$, where $p$ and $q$ are positive integers, $p < q$, then we can write
$$x = x \cdot 1 = \frac{p}{q} \left( \frac{1}{1 \cdot 2 \ldots p} + \frac{1}{2 \cdot 3 \ldots p} + \frac{2}{3 \cdot 4 \ldots p} + \frac{p-1}{p} \right),$$
and we can consider $n = p$, $a_{1} = 1$, $a_{2} = 2$, $\ldots$, $a_{p-1} = p-1$, $a_{p} = q$.

The desired numbers are the elements of the set $\mathbb{Q} \cap (0, 1]$.