The largest three-digit multiple of 9 is 999.
We want the sum of the digits to be 18.
The digits of 999 add up to $9+9+9=27$, which is larger than 18.
To make the sum smaller, we can decrease the hundreds digit by 1 and increase the units digit by 1.
So, the largest three-digit multiple of 9 whose digits' sum is 18 is $\boxed{990}$.
The answer is: 990
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
What is the largest three-digit multiple of 9 whose digits' sum is 18?

### Answer:
