So in this question i first figured out that the numerator must be greater than $5000$ so that there are $4$ digits on the numerator and $5$ digits on the denominator.
And then after a few random guesses:
$$a=9273$$
$$b=18546$$
satisfied the condition.
But i have 2 follow up questions:
On
Well, your denominator has to be of the form 1xxxx, and the last digit of the denominator has to be even. The first digit of the numerator has to be (strictly) greater than 5, since otherwise the denominator will have the form 11... or 10... I am afraid that after that it becomes a search.
On
Just some arithmetical methods to find good bounds on $a$: we know that $\text{the number of digits of the numerator} + \text{number of digits of the denominator}=9$. So $\lfloor \log a\rfloor+1+ \lfloor \log b\rfloor+1=9$ and we had: $b=2a$ so: $\lfloor \log a\rfloor+ \lfloor \log 2a\rfloor=7\geq2\lfloor \log a\rfloor$. Thus, $\lfloor \log a\rfloor\leq3.5$ but $\lfloor \log a\rfloor$ is an integer so $\lfloor \log a\rfloor\leq 3$ and $\log a<4$ and $a\leq10000$ and we should use each non-zero digit once, and the biggest number with that conditions lesser than $10000$ is $9876$. And as you found out, $a>5000$ so with the same method we have: $ 5123<a<9876$.Using some number-theory methods, I think you can find the exact values. I hope this was useful.
Mathematica found $12$ solutions: