Why do all multiples of 99 have a digit sum $\geq 18$? I noticed that this seems to be the case while looking at some multiples.
Q: Can someone come up with a positive conterexample or show that there can't be one?
 A: Hint: The sum of the digits must be a multiple of $9$. The alternating sum of the digits must be divisible by $11$. If the alternating sum is non-zero, then the sum must be greater than $11$, and thus at least $18$...
If the alternating sum is $0$, then can the sum be $9$? 
A: Because in base 10, nonzero multiples of 9 have to have a digit sum that is a multiple of 9, and multiples of 11 must have an alternating sum of digits equal to a multiple of 11. These two things I've just mentioned are well-accepted facts and it sounds like you've done plenty of verification with individual examples.
Let's call $s_1$ the sum of the odd-place digits and $s_2$ the sum of the even-place digits. You're looking to solve $s_1 + s_2 = 9$ and $s_1 - s_2 = 11m$ simultaneously, where $m$ is some integer. That is the only constraint on $m$. And either $s_1$ or $s_2$ may be 0 (not both), but neither may be negative. Furthermore, $s_1 \leq 9$, and likewise for $s_2$. With these constraints, the two equations have no common solution, and therefore no counterexample can exist.
