4. Place 27 balls numbered $1 \sim 27$ into three bowls, Jia, Yi, and Bing, such that the average values of the ball numbers in bowls Jia, Yi, and Bing are $15$, $3$, and $18$, respectively, and each bowl must contain no fewer than 4 balls. Then the maximum value of the smallest ball number in bowl Jia is $\qquad$
4. 10 .

Let there be $a$, $b$, and $c$ balls in bowls 甲, 乙, and 丙, respectively. Then,
$$
\begin{aligned}
a+b+c & =27, \\
15 a+3 b+18 c & =\frac{27 \times 28}{2} .
\end{aligned}
$$

From equation (2), we get
$$
5 a+b+6 c=126 \text {. }
$$

If $b \geqslant 6$, then the average value of the ball numbers in bowl 乙 is not less than
$$
\frac{1+2+\cdots+b}{b}=\frac{b+1}{2}>3,
$$

which is a contradiction.
Therefore, $b \leqslant 5 \Rightarrow b=4$ or 5.
$6 \times$ (1) - (3) gives $a+5 b=36$.
Solving this, we get $(a, b)=(11,5),(16,4)$.
We discuss two cases.
(1) $(a, b, c)=(11,5,11)$.
In this case, bowl 乙 must contain balls numbered 1 to 5. When the ball numbers in bowl 甲 are $10 \sim 20$, the minimum ball number reaches its maximum value of 10.
(2) $(a, b, c)=(16,4,7)$.
In this case, bowl 乙 must contain balls numbered $(1,2,4,5)$ or $(1,2,3,6)$. When the ball numbers in bowl 甲 are $7 \sim 14,16 \sim 23$, the minimum ball number reaches its maximum value of 7.
In summary, the maximum value of the minimum ball number in bowl 甲 is 10.