The question is as stated in the title. I feel like the question is geared towards some kind of case that yields to the divisibility test for $9$, but I think an argument can be made for other numbers as well. For example in the sequence $100$ ,...,$117$ we have of course $3 \mid 102$ and $9 \mid 117$, but we also have $2 \mid 110$ and $4 \mid 112$. And secondly in sequences where the digit sum is above $9$, we still have divisors that are multiples of $9$. For example $18 \mid 990$.
So if anyone could help me with a proof, I'd be really grateful. Thank you for your help.
We can use your idea: there is at least one multiple of $9$ in this sequence, so the sum of its digits will be $9$, $18$ or $27$. Let's separate some cases:
The sum is $27$: in this case, the number is $999$, so $990$ is also in the sequence and satisfies the properties.
The sum is $9$, and there is nothing to do.
The sum is $18$ and the number is even, and there is nothing to do again.
The sum is $18$ and the number is odd: in this case, let $x$ be this multiple of $9$. Then either $x+9$ or $x-9$ will be in the sequence, and it will fall in case 2. or 3. above. Anyway we are done.