4. The four-digit number $\overline{a b c d}$ is divisible by 3, and $a, b, c$ are permutations of three consecutive integers. Then the number of such four-digit numbers is
$\qquad$ .
4. 184 Detailed Explanation: Let $a, b, c$ be a permutation of three consecutive integers $k, k+1, k+2$, then $3 \mid a+b+c+d$, i.e., $3 \mid 3k+3+d$, so $d=0,3,6,9$. When $k=0$, $(a, b, c)$ can be one of the following: $(1,0,2),(1,2,0),(2,0,1),(2,1,0)$.

In these four cases; when $k=1,2,3,4,5,6,7$, $a, b, c$ can be any permutation of $k, k+1, k+2$, with each category having $A_{3}^{3}=6$ cases, so $a, b, c$ have a total of $4+6 \times 7=46$ (cases) of selection. By the multiplication principle, such four-digit numbers total $46 \times 4=184$ (numbers).