Counting the number of 4-digit numbers and sum 9 can be done using stars and bars really fast (this works because $9$ is less than $10$).
There are $9-1=8$ stars (the first number is at least $1$) and $3$ bars. Hence there are $\binom{8+3}{3}=\frac{11\cdot10\cdot9}{3\cdot2}=11\cdot5\cdot3=165$ such numbers.
Now we have to substract the ones that are multiples of $11$. There are none as you have already realized, that is because $(a+c)-(b+d)\neq 0$ as $a+c$ has a different parity as $b+d$ (they add $9$).
I checked with my computer just to make sure it is correct.
Here is the c++ code, it gives $165$.
#include <cstdio>
#include <cstdlib>
int sumd(int x){
int a=0;
while(x!=0){
a+=x%10;
x/=10;
}
return(a);
}
int main(){
int a,b=0;
for(a=1000;a<10000;a++){
if(sumd(a)==9){
b++;
}
}
printf("%d\n",b);
}
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
Hint:
The unused digits sum to $9$.