Subtraction = Digit Sum

Let $A=\overline{ab}$ and $B=\overline{cd}$ , where $a, b, c, d$ are integers and $0 \le a, b, c, d \le 9$

Therefore, $A-B=\overline{ab}-\overline{cd}=(10a+b)-(10c+d)=10(a-c)+(b-d)$

Thus, $a+b+c+d=10(a-c)+(b-d)$

Now realize that $0 \le a+b+c+d \le 36$ and $min(b-d)=-9$ , therefore $0 \le 10(a-c)+(b-d) \le 36$ $0 \le 10(a-c)-9 \le 36$ $0 \le 10(a-c) \le 45$

Since $a$ and $c$ are integers, and also $a-c$ , so $0 \le 10(a-c) \le 40$ $0 \le a-c \le 4$

Now we will consider 5 cases.

Case 1: $a-c=0$

$a+b+c+d=b-d \Rightarrow c+b+c+d=b-d \Rightarrow 2c=-2d \Rightarrow c=-d \Rightarrow c=d=0 \Rightarrow a=c=d=0, 0 \le b \le 9$

Therefore, there are 10 possibilities in this case

We will deal with the cases which $a-c=1, 2, 3, 4$ with the same manner (expressing $a$ in terms of $c$ and substitution), and we end up with the following results:

Case 2: $a-c=1$

$c+d=4.5 \neq integer$

Therefore, an impossible case

Case 3: $a-c=2$

$c+d=9$ and $a-c=2$ and $0 \le b \le 9$

Since $c$ is non-negative, $a \ge 2$ , and we can find 8 triple pair of $(a, c, d)$ with $a$ start from 2 to 9.

For each triple pair $(a, c, d)$ , any $0 \le b \le 9$ is a solution. Thus, we have $8 \times 10 = 80$ possibilities in this case.

Case 4: $a-c=3$

Similar with Case 2, we end up with $c+d=13.5$ , which means another impossible case

Case 5: $a-c=4$

We end up with $c+d=18$ , which means $c=d=9$ and $a=13 \ge 9$ Another impossible case.

Overall, only Case 1 and 3 yields possibilities to the problem, and we get total $10+80=\boxed{90}$ possibilities!