Answer: $48651$.

Suppose that both numbers contain $5$ digits. A common divisor of two different numbers cannot exceed half of the larger one; thus the greatest common divisor of two such numbers must be less than $50000$. It means that if the greatest common divisor has $5$ digits then the first digit is at most $4$. Let the greatest common divisor be $d$ and suppose that it has $5$ digits, the first of which is $4$. Then the two numbers under consideration are $d$ and $2d$. If the second digit of $d$ were $9$ then the first two digits of $2d$ would be either $98$ or $99$, but in both cases, $9$ would occur in these numbers repeatedly. Thus the second digit of $d$ is at most $8$; suppose that it is $8$. Then the first digit of $2d$ is $9$. If the third digit of $d$ were $7$ then the second digit of $2d$ would be $7$, too. Thus the third digit of $d$ is at most $6$ and the fourth digit of $d$ is at most $5$; suppose that the third and the fourth digit are $6$ and $5$, respectively. Then the second and the third digit of $2d$ are $7$ and $3$, respectively, and the fourth digit is $0$ as the remaining digits $1$ and $2$ cannot cause a carry. Finally, $1$ and $2$ must be the last digits of $d$ and $2d$, respectively, yielding $d = 48651$ and $2d = 97302$. These numbers satisfy the conditions of the problem.

If the given integers are not $5$-digit numbers then the greatest common divisor can have at most $4$ digits. Consequently, there cannot be solutions greater than that found in the previous paragraph.