7. The sum of 100 positive integers is 101101. What is the maximum possible value of their greatest common divisor? Prove your conclusion.
7. Solution: Let the greatest common divisor of 100 positive integers $a_{1}, a_{2}, \cdots, a_{100}$ be $d$, and let $a_{j}=d a^{\prime}{ }_{j}(1 \leqslant j \leqslant 100)$. Since $a_{1}+a_{2}+\cdots+a_{100}=d\left(a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a_{100}\right)$
$$
\begin{array}{l}
=101101 \\
=101 \times 1001
\end{array}
$$

Thus, $a_{1}^{\prime}, a_{2}^{\prime}, \cdots, a_{100}^{\prime}$ cannot all be 1, so $a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a^{\prime}{ }_{100} \geqslant 1 \times 99+2=101$.
Therefore, $d \leqslant 1001$.
On the other hand, if we take $a_{1}=a_{2}=\cdots=a_{99}=1001, a_{100}=2002$, then $a_{1}+a_{2}+\cdots+a_{100}=$ 101101.
And $(1001,1001, \cdots, 1001,2002)=1001$,
Thus, the maximum possible value of the greatest common divisor of $a_{1}, a_{2}, \cdots, a_{100}$ is 1001.