Each of $A, B, C$, and $D$ is a positive two-digit integer. These integers satisfy each of the equations

$$
\begin{aligned}
B & =3 C \\
D & =2 B-C \\
A & =B+D
\end{aligned}
$$

What is the largest possible value of $A+B+C+D$ ?
Substituting $B=3 C$ into $D=2 B-C$ gives $D=2(3 C)-C=5 C$.

Substituting $D=5 C$ and $B=3 C$ into $A=B+D$ gives $A=3 C+5 C=8 C$.

Since $A=8 C$ is a two-digit integer, we must have $10 \leq 8 C \leq 99$, so $\frac{10}{8} \leq C \leq \frac{99}{8}$.

Since $C$ is an integer, this implies $2 \leq C \leq 12$.

We also know that $C$ is a two-digit integer, so $10 \leq C$, which means that $C=10, C=11$, or $C=12$.

Using the equations from earlier, we have $A+B+C+D=8 C+3 C+C+5 C=17 C$, so the larger $C$ is, the larger $A+B+C+D$ is.

With $C=12$, we get $A=8(12)=96, B=3(12)=36$, and $D=5(12)=60$, which are all two-digit integers.

Therefore, the conditions are satisfied when $C=12$, its largest possible value, which means $C=12$ corresponds to the largest possible value of $A+B+C+D$.

Therefore, the answer is $A+B+C+D=17 C=17(12)=204$.

ANSWER: 204