18. (10 points) Two natural numbers, the difference between them is 11, and the sum of the digits of each number is divisible by 11. What is the smaller of the two smallest natural numbers that meet the requirements?
【Solution】Solution: Let the "sum of digits" of the smaller number $a$ be $11n$. Then, the "sum of digits" of $a+11$ is $11n+2-9k=(11n-11k)+2+2k=(11n-11k)+2(1+k)$. Therefore, the smallest $k=10$. That is to say, the required $a$ should result in 10 carries when $a+11$ is calculated. To cause a carry, the unit digit of $a$ should be 9, and the tens digit of $a$ should be 9 or 8. To find the smallest $a$, we take the tens digit of $a$ to be 9.

Furthermore, to make the "sum of digits" of $a$ a multiple of 11, the hundreds, thousands, and so on, up to a certain place, should all be 9, until there are 10 carries, and then add appropriate numbers in front to ensure: no further carries, the "sum of digits" is a multiple of 11, and $a$ is the smallest.

Thus, it should be $a=189999999999$, and the other number is $189999999999+11=190000000010$.