4. For the three-digit number $\overline{a b c}$, satisfying $\overline{a b c}=37(a+b+c)$, the number of such three-digit numbers is ( ) .
(A) 13
(B) 14
(C) 15
(D) 16
4. C.

From the problem, we know
$$
\begin{array}{l}
100 a+10 b+c=37(a+b+c) \\
\Leftrightarrow 7 a=3 b+4 c \Leftrightarrow 4(a-c)=3(b-a) .
\end{array}
$$

Therefore, when $a=b=c$, there are 9 cases;
When $a-c=3, b-a=4$,
$$
b-c=7 \Rightarrow \overline{a b c}=370,481,592 \text {; }
$$

When $a-c=-3, b-a=-4$,
$$
b-c=-7 \Rightarrow \overline{a b c}=407,518,629 \text {. }
$$

In summary, there are 15 three-digit numbers that satisfy $\overline{a b c}=37(a+b+c)$.