Let's determine the smallest positive number divisible by 999 that does not contain the digit 9.
We are looking for the smallest positive integer $A$ such that

$$
B=999 A=1000 A-A
$$

does not contain the digit 9 in $A$. From both forms of $B$, it is clear that the last digit of $A$ cannot be 1, since in that case the last digit of $B$ would be 9. The last digit of $A$ cannot be 0 either, because then a tenth of $B$ would also be a valid number. From the second form of $B$, it is clear that for this reason, the second last digit of $A$ cannot be 0 either, because then performing the subtraction $1000 A-A$ digit by digit in the usual way, we would get a 9 as the second last digit of $B$, regardless of the other digits of $A$. However, this means that the digit in the hundreds place of $A$ cannot be 0 either, because then the digit in the hundreds place of $B$ would be 9. If, therefore, we choose every digit of $A$ to be as small as possible, we get the number 112, and with it $B=111888$ does not contain a 9, so this is the number we are looking for.

Remark. Similarly, it can be seen that if in the problem we use the number written with $k$ nines instead of 999, the number we are looking for consists of $k$ ones followed by $k$ eights.