Multiples of $999$ have digit sum $\geq 27$

Lemma. Let $n$ be an integer $\ge 1000$. Then there exists a positive integer $m$ such that $m<n$, $n-m$ is a multiple of $999$ and for the decimal digit sums, we have $q(m)\le q(n)$.

Proof. $n$ has a $k$-digit decimal expansion $n=\overline{a_ka_{k-1}\ldots a_1}$ (with $k\ge 4$ and $a_k\ge1$), then $m:=n-999\cdot 10^{k-4}$ is non-negative and has a decimal expansion $m=\overline{b_kb_{k-1}\ldots b_1}$, where $b_j=a_j$ for all $j$ except $$\begin{cases}b_k=a_k-1,b_{k-3}=a_{k-3}+1&\text{if }a_{k-3}<9\\ b_k=a_k-1,b_{k-3}=0, b_{k-2}=a_{k-2}+1&\text{if }a_{k-2}<a_{k-3}=9\\ b_k=a_k-1,b_{k-2}=b_{k-3}=0, b_{k-1}=a_{k-1}+1&\text{if }a_{k-1}<a_{k-2}=a_{k-3}=9\\ b_{k-1}=b_{k-2}=b_{k-3}=0&\text{if }a_{k-1}=a_{k-2}=a_{k-3}=9\\ \end{cases} $$ Then for the digit sum of $m$ we find accordingly $$q(m)=\begin{cases}q(n)\\q(n)-9\\q(n)-18\\q(n)-27\end{cases}\le q(n) $$ Hence if $m>0$, the claim follows. On the other hand, if $m=0$, it follows that $n=999\cdot 10^{k-4}$, $q(n)=27$, and we can take $m=999$. $\square$

Corollary. If $n$ is a positive multiple of $999$, then $q(n)\ge 27$.

Proof. By the lemma, the set of positive multiples of $999$ with digit sum $<27$ has no smallest element. $\square$

Just a partial answer

This is true for all 3 digit $k$.

Let $k=\overline{abc}$.

The $999k=\overline{abc000}-abc$.

When $c\ne0$:

For the difference:

Unit digit is $10-c$.

Tens digit is $9-b$.

Hundreds digit is $9-a$.

Thousands digit is $c-1$.

Ten thousands digit(?) is $b$.

Hundred thousands digit(?) is $a$.

Thus the sum of digit is exactly $27$.

---

A similar approach can prove for the case $c=0, b>0$ and $c=0,b=0$.

Let us consider some examples, all steps in the argumentation are then also applied on the examples:

3300652000033011
12345678987654321

(1) We start with a number written in base $10$, which is divisible by $999$. We break it in blocks of numbers of three digits, starting from the units digit, where we find the "first block". The last block may be incomplete", in this case we may add or not zeros in front of it. Because $1000$ is congruent to one modulo $999$, the sum of these blocks, considered as numbers between $0$ and $999$, is also divisible by $999$.

In our case, we separate the groups

3.300.652.000.033.011
12.345.678.987.654.321

obtain the blocks

003 and respectively 012
300                  345
652                  678
000                  987
033                  654
011                  321

and the sum of the corresponding numbers is $999$, and respectively $2997$. It stays divisible by $999$. We want to show that the sum of the digits of the numbers in the blocks is at least $27$.

(2) We repeat this operation till we get a number of three digits. This number is of course $999$ in the first case. In the second one we group again 002 and 997, add, get $999$, and stop here.

(3) To finish the proof we note the fact that looking at the sum of the digits in the "blocks" before and after applying the step (1), the sum drops (by a multiple of $9$), it was before bigger than after. This has something to do with the algorithm we learn first in school. We put two numbers over each other. We add the unit digits. If the result is $\le 9$, then the contribution of the digits to the sum of digits of the two numbers we start with is the same as the corresponding contribution in the result. Else we have a drop by $9$. This goes forward for the next digits...

Inductively we are done.

Note: There is "nothing special" about $999$, compared to $9$, $99$, ... , $\underbrace{99\dots99}_{n\text{ digits}}$, the same works by building blocks of length $n$ (in the general case, the last one explicitly listed).