1. Given the sum of 12 distinct positive integers is 2010. Then the maximum value of the greatest common divisor of these positive integers is . $\qquad$
$-1.15$.
Let the greatest common divisor be $d$, and the 12 numbers be $a_{1} d$, $a_{2} d, \cdots, a_{12} d$, where $\left(a_{1}, a_{2}, \cdots, a_{12}\right)=1$.
Let $S=\sum_{i=1}^{12} a_{i}$. Then $2010=S d$.
To maximize $d$, $S$ should be minimized.
Since $a_{1}, a_{2}, \cdots, a_{12}$ are distinct, then
$S \geqslant 1+2+\cdots+12=78$.
Also, $S \mid 2010$, and $2010=2 \times 3 \times 5 \times 67$, so the smallest positive divisor of 2010 greater than 77 is $2 \times 67=134$.
Therefore, $d \leqslant 15$, and $d=15$ can be achieved, by setting
$$
\left(a_{1}, a_{2}, \cdots, a_{11}, a_{12}\right)=(1,2, \cdots, 11,68)
$$

This completes the translation.