$5 \cdot 4$ natural numbers $a_{1}, a_{2}, \cdots, a_{49}$ have a sum of 999, let $d$ be the greatest common divisor of $a_{1}, a_{2}, \cdots, a_{49}$, what is the maximum value of $d$?
(Kiev Mathematical Olympiad, 1979)
[Solution] Since $d=\left(a_{1}, a_{2}, \cdots, a_{49}\right)$, then
$$
d \mid a_{1}+a_{2}+\cdots+a_{49}=999 \text {. }
$$

That is, $d$ is a divisor of $999=3^{3} \cdot 37$.
Also, because
then
$$
\begin{aligned}
d \mid a_{k}, \quad k & =1,2, \cdots, 49, \\
a_{k} & \geqslant d .
\end{aligned}
$$

Therefore, there must be $999=a_{1}+a_{2}+\cdots+a_{49} \geqslant 49 d$,
$$
d \leqslant \frac{999}{49}<21 \text {. }
$$

Furthermore, since $d$ is a divisor of 999, $d$ can only be $1,3,9$.
Also, because 999 can be written as the sum of 49 numbers.
$$
\underbrace{9+9+\cdots+9}_{48 \uparrow}+567=999 .
$$

Each of these numbers can be divided by 9, so the maximum value of $d$ is 9.