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Let $N=\overline{abcd}$ . Recall that $N\equiv a + b + c + d\bmod 9$ . Since $4N\equiv d + c + b + a \bmod 9$ , it must be that $4N \equiv N\bmod 9$ . This in turn implies that one of the following must be true $\begin{aligned} N&\equiv 0\mod 9 \\ N&\equiv 3\mod 9 \\ N&\equiv 6\mod 9 \end{aligned}$ In all cases, $N$ is divisible by $3$ . Thus the digital sum is also divisible by 3. 18 is the only answer that is divisible by 3.

We are given

\(\begin{array} {} & a & b & c & d \\ \times & & & & 4 \\ \hline & d & c & b & a \end{array} \)

Since $\overline{dcba}$ is a multiple of 4, $a$ can only be 0, 2, 4, 6, 8. Obviously $a < 3$ , else $\overline{abcd} \times 4$ would have 5 digits. Since $a \ne 0$ , $a$ must be 2. Now the leading $d$ of $\overline{dcba}$ can only be 8 or 9 depending if there is a carry-forward of 1 from behind. But it is obvious that $d = 8$ because $8 \times 4 = 32$ where the end 2 is $a$ . And we have:

\(\begin{array} {} & 2 & b & c & 8 \\ \times & & & _{\color{red}3} & 4 \\ \hline & 8 & c & b & 2 \end{array} \), where the red $\color{#D61F06}3$ is a carry-forward.

From the third column, we get $4c + 3 \equiv b \text{ (mod 10)}$ , this means that $b$ is odd. Again $b < 3$ , else the leading $d$ of $\overline{dcba}$ is not 8. So $b=1$ and $4c + 3 \equiv 1 \text{ (mod 10)}$ $\implies c =$ 2 or 7. And $c=7$ fits the bill because:

\(\begin{array} {} & 2 & 1 & 7 & 8 \\ \times & & _{\color{red}3} & _{\color{red}3} & 4 \\ \hline & 8 & 7 & 1 & 2 \end{array} \)

Therefore, $a+b+c+d = 2+1+7+8 = \boxed{18}$ .