Digit sum of 99

[This is not yet a complete solution, but has enough for you to work it out.]

The sum of digits has to be a multiple of 9.
The alternating sum of digits has to be a multiple of 11.

$99 \times 1= 99 \ and \ 9+9=18$ . Talking about multiples is another way of looking at the multiplication table of $99$ , then the smallest multiple will the one where $99$ multiplied by $1$ .

My solution is about the properties of multiples of 11:

When you multiply a number by $11$ , you sum the digits of the number 2 by 2. For example:

Multiplying $24\times11$ you get $264$ as answer. In fact $2+4=6$ . Multiplying $38\times11$ you get $418$ as answer. In fact $3+8=11$ and carrying 1, $3+1=4$ .

Now returning to the main problem.

When you multiply every number by 99, you are multiplying this number by 9 and 11.

The sum of digits of a multiple of 9 is always a multiple of 9.

Let's try with number 7:

$99\times7=7\times9\times11=63\times11$ . In this case the numbers 6 and 3 sum up to 9. Then the answer will be $693$ . In this case the sum of digits is 18. And it will continue until 100, when you get 9900.

After this you will get two more 9's in the sum, and the digit sum will be 36, when you multiply by 101. Other numbers after it will sum or 18 or higher numbers. It is because you will always have a 9 in the answer or two or more groups of digits that sum up to 9. Then 9 is not the minimum.

The proof of it is because the alternating sum of digits of a multiple of 11 is a multiple of 11. It means that a number in form $\text{abcdef...}$ is a multiple of 11 if $(a+c+e+...)-(b+d+f+...)$ is too. If these digits sum up to 9, it's impossible to get a multiple of 11, because the alternating sum must be a multiple of 11. Because 9 is an odd number, the values for the sum of digits in even position ( $(a+c+e+...)$ ) and odd position ( $(b+d+f+...)$ ) must be 9 and 0, 8 and 1, 7 and 6, 5 and 4, 4 and 5, 3 and 6, 2 and 7, 1 and 8 or 0 and 9. Making a subtraction of it you will not get a multiple of 11.

We can conclude about it, is that the sum of digits of a multiple of 99 will never be smaller than 18. It is because the properties of multiplying by 11 and 9.