What is the smallest positive integer $n$, such that there exist positive integers $a$ and $b$…

What is the smallest positive integer $$n$$, such that there exist positive integers $$a$$ and $$b$$, with $$b$$ obtained from $$a$$ by a rearrangement of its digits, so that $$a – b = 11\dots 1$$ (The number of '$$1$$'s equal to $$n$$)?

Is there any example that can satisfy the equation?

Hint 1. Since $$b$$ is obtained as a rearrangement of the digits of $$a$$, it follows that $$a$$ and $$b$$ have the same remainder when they are divided by $$9$$. Hence $$a-b$$ should be divisible by $$9$$.
Hint 2. Note that $$90-09=81,\ 190-019=171,\ 1290-0129=1161,\ 12390-01239=11151, \dots$$
Since $$b = \text{DigitsRearranged}(a) \rightarrow$$ $$a$$ and $$b$$ have the same remainder when divided by $$9$$ $$\rightarrow$$ $$a-b$$ is divisible by $$9$$ $$\rightarrow$$ $$n \ge 9$$
And this is sufficient, because there is a solution for this $$n$$: $$a = 812,345,709 \\ b = 701,234,598 \\ a-b = 111,111,111$$
So the smallest $$n$$ is: $$n = 9$$.