The number has to be $15317$. I give in to the temptation of posting my earlier comment as an answer. The number ends in $17$, so we have a starting point, the digit sum is $8$. We now need a digit sum of $9$, and the number appended on the left has to be divisible by $9$. Since $17$ and $9$ are coprime, $17$ x $9$ = $153$ is the smallest such number. Hence the answer to the original question is $15317$.
As you note, the requirement for a digit sum of $43$ restricts the possible digit choices quite sharply.
In general numbers divisible by $11$ have alternating digit sums (eg. adding $a$s and $b$s from $abababa$) that differ by a multiple of $11$ (which could be zero, if sums are equal). This is due to powers of $10$ alternating between $-1$ and $1 \bmod 11$.
In this case since the total digit sum is odd, we must have alternating sums that differ by $11$, that is $16$ and $27$ for the two- and three-digit sums respectively. Clearly this gives us something of the form $9\;\square\;9\;\square\;9$ and the options for the intervening digits adding to $16$ are $(7,9),(8,8),(9,7)$ as you found.
By request in comments:
If the digit sum were $36$, all other conditions unchanged, we would have to have the digit sums equal $\to (18,18)$ (as a difference of $22 \to (29,7)$ is not feasible). Then there is only one option for the two-digit set $(9,9)$ and we can find the number of divisions on the three-digit set by a small inclusion-exclusion to partition the $18$ into three valid digits. First there are two options that include a zero, $(99099,99990)$ after which we can take the partitions as being non-zero. Then without constraint on the upper size of partition, the options would be $\binom {17}2$, and removing the cases with a digit greater than $9$ reduces this by $3\cdot \binom 82$, giving $2+136-84=54$ options.
Best Answer
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 $N$.