1. The sum of 100 positive integers is 101101. What is the maximum possible value of their greatest common divisor? Prove your conclusion.
1. Let the greatest common divisor of 100 positive integers $a_{1}, a_{2}, \cdots, a_{100}$ be $d$, and let $a_{j}=d a_{j}^{\prime}(1 \leqslant j \leqslant 100)$.
Since $a_{1}+a_{2}+\cdots+a_{100}=d\left(a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a_{100}^{\prime}\right)=101101=101 \cdot 1001$, it follows that $a_{1}^{\prime}, a_{2}^{\prime}, \cdots, a_{100}^{\prime}$ cannot all be 1, thus
$$
a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a_{100}^{\prime} \geqslant 1 \cdot 99+2=101 \text {. }
$$

Therefore, $d \leqslant 1001$.
On the other hand, if we take $a_{1}=a_{2}=\cdots=a_{99}=1001, a_{100}=2002$, then it satisfies $a_{1}+a_{2}+\cdots+a_{100}=101101$.
And $(1001,1001, \cdots, 1001,2002)=1001$,
so the maximum possible value of the greatest common divisor of $a_{1}, a_{2}, \cdots, a_{100}$ is 1001.