Problem: A rational number $x$ is given.
Prove that there exists a sequence
$x_0, x_1, x_2, \dots$ of rational numbers
with the following properties:
\begin{enumerate}[(a)]
  \item $x_0=x$;
  \item for every $n\ge1$, either $x_n = 2x_{n-1}$ or $x_n = 2x_{n-1} + \frac{1}{n}$;
  \item $x_n$ is an integer for some $n$.
\end{enumerate}

Solution: Think of the sequence as a process over time.
We'll show that:
\begin{claim*}
  At any given time $t$,
  if the denominator of $x_t$ has some odd prime power $q = p^e$,
  then we can delete a factor of $p$ from the denominator,
  while only adding powers of two to the denominator.
\end{claim*}
(Thus we can just delete off all the odd primes one by one
and then double appropriately many times.)

\begin{proof}
The idea is to add only fractions of the form $(2^k q)\inv$.

Indeed, let $n$ be large,
and suppose $t < 2^{r+1} q < 2^{r+2} q < \dots < 2^{r+m} q < n$.
For some binary variables $\eps_i \in \{0,1\}$ we can have
\[ x_n = 2^{n-t} x_t
  + c_1 \cdot \frac{\eps_1}{q}
  + c_2 \cdot \frac{\eps_2}{q}
  \dots
  + c_s \cdot \frac{\eps_m}{q}
\]
where $c_i$ is some power of $2$
(to be exact, $c_i = \frac{2^{n-2^{r+i}q}}{2^{r+1}}$,
but the exact value doesn't matter).

If $m$ is large enough the set $\{0,c_1\} + \{0,c_2\} + \dots + \{0, c_m\}$
spans everything modulo $p$.
(Actually, Cauchy-Davenport implies $m=p$ is enough,
but one can also just use Pigeonhole to notice some residue appears
more than $p$ times, for $m = O(p^2)$.)
Thus we can eliminate one factor of $p$ from the denominator, as desired.
\end{proof}