Cryptogram.

$\small \begin{aligned} \overline{ABCDC} \times 9 & = \overline{CDCBA} \\ 90000A + 9000B + 900C + 90D + C & = 10000C + 1000D + 100C + 10B + A \\ 89999A + 8990B & = 9191C + 910D = 91(101C +10D) \end{aligned}$

We note that the RHS is divisible by 91, therefore the LHS is also divisible by 91 or:

$\begin{aligned} 89999A + 8990B & \equiv 0 \text{ (mod 91)} \\ (0)A +72B & \equiv 0 \text{ (mod 91)} \\ \implies B & = 0 \end{aligned}$

Then we have:

$\begin{aligned} 89999A & = 9191C + 910D & \small \blue{\text{Divide both sides by }91} \\ 989 A & = 101C + 10D & \small \blue{\text{RHS is maximum when }C=D=9} \\ 989 A & \le 999 & \small \blue{\text{For }A \ne 0} \\ \implies A & = 1 \\ \implies 989 & = 101C + 10D = \overline{CDC} \\ \implies C & = 9 \implies D = 8 \end{aligned}$

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

Since the multiplication product is a five digit number, $A$ must be $1$ and $B$ must be $0$ . Hence $C=9\times 1=9$ and $D=8$ . Therefore $A+B+C+D=\boxed {18}$