Solution:

The greatest common divisor of two numbers also divides their difference. If $A$ and $B$ are two distinct digits of $n$ with $A>B$, the two numbers $XXXAB$ and $XXXBA$ are in $A_{n}$ ($XXX$ represents any sequence of the other digits of $n$). Their difference is $9(A-B)$.

The maximum value of $A-B$ is $9$, which is achieved only if $A=9$ and $B=0$. This shows that $d_{n}$ is less than or equal to $81$. Let us now see that there exists an integer $N$ such that $d_{N}=81$. Such an $N$ must be formed only by digits $9$ and $0$ (otherwise one finds $A$ and $B$ as above, with $A-B<9$, while $d_{N} \leq 9(A-B)<81$) and must itself be a multiple of $81$. Equivalently $N/9$ must be formed only by digits $1$ and $0$ and must be a multiple of $9$. By the divisibility rule for $9$, the sum of the digits of $N/9$ must be a multiple of $9$, and therefore $N/9$ must have at least $9$ digits equal to $1$.

The smallest integer in $X$ with these properties is $1011111111$, so $N$ is at least $9099999999$. On the other hand, if $N=9099999999$, one easily checks that all the numbers of $A_{N}$ are divisible by $81$, and therefore $d_{N}=81$.