First, we all know $\mathsf{A=1, D=9}$ .

$\overline{\mathsf{1BC9}} = 1009 + \overline{\mathsf{BC}} \times 10$ .

$\overline{\mathsf{1BC9}} \times 9 = 9081 + \mathsf{B} \times 900 + \mathsf{C} \times 90$ .

$\overline{\mathsf{9CB1}} = 9001 + \mathsf{B} \times 10 + \mathsf{C} \times 100$ .

Now we have this equation:

$9081 + \mathsf{B} \times 900 + \mathsf{C} \times 90 = 9001 + \mathsf{B} \times 10 + \mathsf{C} \times 100$ .

$\mathsf{B} \times 890 + 80 = \mathsf{C} \times 10$ .

$\mathsf{B} \times 89 + 8 = \mathsf{C}$ .

If $\mathsf{C} \not = 8$ , there's no solution for $\mathsf{B}$ .

If $\mathsf{C} = 8$ then $\mathsf{B} = 0$ .

$\mathsf{A + B + C + D} = 1 + 0 + 8 + 9 = \boxed{18}$ .

Here's a step by step solution:

$\frac{\begin{array}{c}[b]{r} ABCD\\ \times\quad\quad\quad9 \end{array}} {\begin{array}{c}[b]{r} \quad\;\; DCBA\\ \hline \end{array}}$

Now, we know that $A \neq 0$ . If $A>1$ , the product will be a 5 digit number, and that does not satisfy the cryptogram. From here, we can confirm that $A = 1$

From here, we know that $D \times 9$ gives us a number that ends with $1$ . Looking at the multiplication table, only $9 \times 9 = 81$ satisfies the requirement. Therefore,

$D = 9$

Substitute these two numbers into the cryptogram, and we get:

$\frac{\begin{array}{c}[b]{r} 1BC9\\ \times\quad\quad\quad9 \end{array}} {\begin{array}{c}[b]{r} \quad\quad\; 9CB1\\ \hline \end{array}}$

Now, we convert it into an equation as below:

$(1009 + 100B + 10C)(9) = 9001 + 100C + 10B$

Simplifying this equation gives:

$89B + 8 = C$

Now, remember that the values of $B$ and $C$ are integers for a digit, which gives:

$0 \leq B \leq 9 \quad\quad 0 \leq C \leq 9$

If $B>0$ , $C$ will definitely be larger than $9$ , and that does not satisfy the cryptogram.

Therefore, the only solution is $B = 0$ , and $C = 89(0) + 8 = 8$ .

The solved cryptogram is

$\frac{\begin{array}{c}[b]{r} 1089\\ \times\quad\quad\quad9 \end{array}} {\begin{array}{c}[b]{r} \quad\quad\;\; 9801\\ \hline \end{array}}$

Calculate the final answer: $A + B + C + D = 1 + 0 + 8 + 9 = \boxed{18}$

If ABCD is greater than 1111, then the solution will be a 5 digit number, and thus be null.

Therefore, A must equal 1, and B 0. Therefore, 1000<ABCD<1111. Any of these values multiplied by 9 give 9 as the leading coefficient.

From here, C is easy to solve for, by doing multiplication we can see that the 8 will carry over from D*9, giving 8+9C = 10C+0, C = 8.

(Sorry, that last bit was poorly worded)

The four digit number when multiplied by 9 gives a 4 digit number which means that A must be 1 and since the result of A * 9=D , D must be 9. Since the thousand digit of the result is the result of A * 9 plus the result of the decimal digit of B * 9 B must be 0 and since D * 9 = 9 * 9=81 it means that the decimal digit of the result that is B which is 0 is the result of 8 to which we add the unit digit of the product C*9 this being 2 therefore being that C is 8 which means that A = 1 , B = 0 , C = 8 , D = 9 which added gives 18.

The numbers are: $\mathsf{A}=1, \mathsf{B}=0, \mathsf{C}=8, \mathsf{D}=9$ .

The cryptarithm was $1089 \times 9 = 9801$ .