For every $N$ there is a number $X$ such that $N$ divides $X$ and the sum of digits of $X$ equals $N$.
Proof: Write $N = RM$ where $M$ is coprime to $10$ and $R$ contains only the prime factors $2$ and $5$. Then, by Euler's theorem, $10^{\varphi(M)} \equiv 1 \pmod M$. Consider X' := \sum_{i=1}^{N} 10^{i\varphi(M)}. It is a sum of $N$ numbers each of which is congruent to $1$ modulo $M$, so X' \equiv N\cdot 1 \equiv 0 \pmod M. Furthermore, the decimal representation of X' contains exactly $N$ ones, all other digits are $0$, so the sum of digits of X' is $N$. Multiply X' by a high power of ten to get a multiple of $R$, call the result $X$. Then $X$ is divisible by $M$ and $R$, hence by $N$, and it has the same digit sum as X' which is $N$.