SAMC

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 $$\begin{array}{rl}& \frac{1}{a_1{a}_2\dots a_n}\le \frac{1}{a_2{a}_3\dots a_n}\\ & \frac{a_1}{a_2{a}_3\dots a_n}\le \frac{a_2-1}{a_2{a}_3\dots a_n}=\frac{1}{a_3{a}_4\dots a_n}-\frac{1}{a_2{a}_3\dots a_n}\\ & \frac{a_{n-2}}{a_{n-1}{a}_n}\le \frac{a_{n-1}-1}{a_{n-1}{a}_n}=\frac{1}{a_n}-\frac{1}{a_{n-1}{a}_n}\\ & \frac{a_{n-1}}{a_n}\le \frac{a_n-1}{a_n}=1-\frac{1}{a_n},\end{array}$$ we get $x\le 1-\frac{1}{a_n}$, 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 $$x=x\cdot 1=\frac{p}{q}\left(\frac{1}{1\cdot 2\dots p}+\frac{1}{2\cdot 3\dots p}+\frac{2}{3\cdot 4\dots p}+\frac{p-1}{p}\right),$$ and we can consider $n=p$, $a_1=1$, $a_2=2$, $\dots$, $a_{p-1}=p-1$, $a_p=q$.

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