Let's find the smallest positive multiple of 45 whose decimal representation consists only of the digits 0 and 8.
The sought number is divisible by 5 and 9, as it is a multiple of 45. Conversely, if a number is divisible by 5 and 9, then due to $(5,9)=1$, it will also be divisible by $5 \cdot 9=45$. Therefore, we need to find the smallest positive integer $K$ whose decimal representation consists only of the digits 0 and 8, and which is divisible by both 5 and 9.

According to the condition, $K$ is even, so it will be a multiple of 5 if its last digit is 0.

A number is exactly divisible by 9 if the sum of its digits has this property. In $K$, only the digit 8 appears besides 0; the smallest multiple of 9 that is a multiple of 8 is $9 \cdot 8$, so the number we are looking for is ten digits long, with the first nine digits being 8 and the last one being $0; K=8888888880$.