4. If the sum of 12 distinct positive integers is 2016, then the maximum value of the greatest common divisor of these positive integers is $\qquad$ .
4. 24 .

Let the greatest common divisor be $d$, and the 12 numbers be $a_{1} d$, $a_{2} d, \cdots, a_{12} d\left(\left(a_{1}, a_{2}, \cdots, a_{12}\right)=1\right)$.
Let $S=\sum_{i=1}^{12} a_{i}$. Then, $2016=S d$.
To maximize $d$, $S$ should be minimized.
Since $a_{1}, a_{2}, \cdots, a_{12}$ are distinct, then
$$
S \geqslant 1+2+\cdots+12=78 \text {. }
$$

Note that, $S \mid 2016$, and 78 is not a factor of 2016.
Also, $2016=2^{5} \times 3^{2} \times 7$, its smallest positive factor greater than 78 is
$$
2^{2} \times 3 \times 7=84,2016=84 \times 24 .
$$

Thus, $d \leqslant 24$, and $d=24$ can be achieved, by setting
$$
\left(a_{1}, a_{2}, \cdots, a_{11}, a_{12}\right)=(1,2, \cdots, 11,18) .
$$