We find that $$D|\overline{ABC}+\overline{BCA}+\overline{CAB}=111(A+B+C)$$and $$D|\overline{(A+1)CB}+\overline{CB(A+1)}+\overline{B(A+1)C}=111(A+B+C)+111$$So, $D$ must be a divisor of $111$. This gives $3$ possible cases. Case I. $D=3$ Note that $\overline{ABC}\equiv A+B+C\pmod3$ and $\overline{(A+1)CB}\equiv A+B+C+1\pmod3$ which means that they cannot both be divisible by $3$ at the same time. Therefore, $D$ cannot be divisible by $3$. Case II. $D=37$ Then, $\overline{(A+1)CB}-\overline{ABC}=100-9B+9C\equiv\pmod{37}\implies B-C\equiv7\pmod{37}$. This gives $(B,C)=(7,0),(8,1),(9,2)$. If $(B,C)=(7,0)$, then $37|\overline{A70}\implies A\equiv3\pmod{37}$ If $(B,C)=(8,1)$, then $37|\overline{A81}\implies A\equiv4\pmod{37}$ If $(B,C)=(9,2)$, then $37|\overline{A92}\implies A\equiv5\pmod{37}$ This gives $(3,7,0,37),(4,8,1,37),(5,9,2,37)$. Case III. $D=111$ Similar to Case I. Thus, the only solutions are $(A,B,C,D)=\boxed{(3,7,0,37),(4,8,1,37),(5,9,2,37)}$.