$5 \cdot 5100$ positive integers sum to 101101. What is the greatest possible value of their greatest common divisor? Prove your conclusion.
(Shanghai Middle School Mathematics Competition, 1986)
[Solution] Let the greatest common divisor of 100 positive integers $a_{1}, a_{2}, \cdots, a_{100}$ be $d$, and let $a_{j}=d a_{j}^{\prime} \quad(1 \leqslant j \leqslant 100)$.

Since
$$
\begin{aligned}
& a_{1}+a_{2}+\cdots+a_{100} \\
= & d\left(a_{1}{ }^{\prime}+a_{2}{ }^{\prime}+\cdots+a^{\prime}{ }_{100}\right) \\
= & 101101 \\
= & 101 \cdot 1001
\end{aligned}
$$

Thus, $a_{1}{ }^{\prime}, a_{2}{ }^{\prime}, \cdots, a^{\prime}{ }_{100}$ cannot all be 1, hence
$$
a_{1}^{\prime}+a_{2}{ }^{\prime}+\cdots+a^{\prime}{ }_{100} \geqslant 1 \cdot 99+2=101 \text {. }
$$

Therefore,
$$
d \leqslant 1001 \text {. }
$$

On the other hand, if we take $a_{1}=a_{2}=\cdots=a_{99}=1001, a_{100}=2002$, then we have
$$
a_{1}+a_{2}+\cdots+a_{100}=101101 .
$$

And $\quad(1001,1001, \cdots, 1001,2002)=1001$.
Therefore, the maximum possible value of the greatest common divisor of $a_{1}, a_{2}, \cdots, a_{100}$ is 1001.