Example 2 Given that the sum of 10 natural numbers is 1001. What is the maximum possible value of their greatest common divisor?

 untranslated text:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 

Note: The last part of the text is a note about the translation instruction and is not part of the content to be translated. Here is the translation with the note removed:

Example 2 Given that the sum of 10 natural numbers is 1001. What is the maximum possible value of their greatest common divisor?
Solution: Let the greatest common divisor be $d$, and the 10 numbers be $b_{1} d, b_{2} d, \cdots, b_{10} d$. According to the problem, we have
$$
\left(b_{1}+b_{2}+\cdots+b_{10}\right) d=1001 .
$$

Since $b_{1}+b_{2}+\cdots+b_{10} \geqslant 10$, it follows that $d \leqslant$ 100 and is a divisor of 1001.

Given $1001=7 \times 11 \times 13$, the largest possible value of the divisor less than 100 is $d=7 \times 13=91$.

When the 10 numbers are $91,91, \cdots, 91,182$, their sum is 1001, so $d=91$ satisfies the condition.