Example 4 Positive integers $a_{1}, a_{2}, \cdots, a_{49}$ 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 $d$?
Analysis From $a_{i} \geqslant d,(i=1,2, \cdots, 49)$ and $d$ is a divisor of 999, we can determine the range of $d$. By checking and eliminating one by one, we can get the $d$ that meets the requirements.
Solution Since $d=\left(a_{1}, a_{2}, \cdots, a_{49}\right)$, then $d \mid\left(a_{1}+a_{2}+\cdots+a_{49}\right)$, i.e., $d \mid 999$,
Therefore, $d$ is a divisor of $999=3^{3} \cdot 37$.
Since $d \mid a_{k}, k=1,2, \cdots, 49$, then $a_{k} \geqslant d$,
Thus, it must be that $999=a_{1}+a_{2}+\cdots+a_{49} \geqslant 49 d$, leading to $d \leqslant \frac{999}{49}<21$.
Furthermore, since $d$ is a divisor of 999, $d$ can only be $1,3,9$.
Also, 999 can be written as the sum of 49 numbers: $\underbrace{9+9+9+\cdots+9}_{48 \text { १ }}+567=999$.
Each of these numbers can be divided by 9, so the maximum value of $d$ is 9.