Which is the smallest positive integer that is divisible by 111 and whose last four digits are 2004?
Let $A$ be the number written before 2004. According to the condition, $A \cdot 10^{4} + 2004$ is divisible by 111:

$$
10000 A + 2004 = 90 \cdot 111 A + 10 A + 18 \cdot 111 + 6
$$

This will be divisible by 111 if $111 \mid 10 A + 6$ holds, where $A$ is at least two digits, since $10 A + 6$ is at least three digits. Among the three-digit numbers, only those are divisible by 111 where all digits are equal. In our case, the last digit being 6, the smallest possibility is $A = 66$. Therefore, the smallest number that satisfies the requirements is: 662004.