Example 7 If the perimeter of a hexagon is 20, each side length is an integer, and no triangle can be formed with any three sides, then such a hexagon ( ).
(A) does not exist
(B) exists uniquely
(C) has a finite number, but more than one
(D) has infinitely many
Analysis: Observing the four options, due to the instability of an $n$-sided polygon $(n \geqslant 4)$, if there exists such a hexagon, there must be infinitely many. Therefore, we need to find whether there exist six positive integers $a_{1}, a_{2}, \cdots, a_{6}$ (without loss of generality, assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{6}$) that satisfy:
1) $a_{1}+a_{2}+\cdots+a_{6}=20 ;$
2) $a_{1} \leqslant a_{2}, a_{1}+a_{2} \leqslant a_{3}, a_{2}+a_{3} \leqslant a_{1}, a_{3}+a_{1} \leqslant a_{5}, a_{4}+a_{5} \leqslant a_{6}$
3) $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}>a_{6}$.

If such a hexagon exists, then the hexagon with side lengths $a_{1}, a_{2}, \cdots, a_{6}$ meets the requirements. In fact, for any three integers $1 \leqslant a_{i} \leqslant a_{j} \leqslant a_{k} \leqslant a_{6}$, it must be that $a_{i}+a_{j} \leqslant a_{k}$, indicating that the three sides cannot form a triangle. For example, if $a_{1}=a_{2}=1, a_{3}=2, a_{4}=3, a_{5}=5, a_{6}=8$, all the conditions are satisfied. Therefore, the answer is D.