Example 4. The perimeter of a hexagon is 20, each side is an integer, and no three sides can form a triangle. Then, such a hexagon ( ).
(A) does not exist.
(B) is unique.
(C) has a finite number,
but more than one.
(D) has an infinite number.
If we can find six integers $a_{1}, a_{2}, \cdots, a^{\theta}$, satisfying: (1) $a_{1}+a_{2}+\cdots+a_{8}=20$; (2) $a_{1} \leqslant a_{2}, \quad a_{1}+a_{2} \leqslant a_{3}, \quad a_{2}+a_{3} \leqslant a_{4}$, $a_{3}+a_{4} \leqslant a_{5}, a_{4}+a_{5} \leqslant a_{63}$ (3) $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}>a_{8}$. Then the hexagon with side lengths $a_{1}, a_{2}, \cdots, a_{8}$ will meet the requirements.

In fact, for any three integers $1 \leqslant i \leqslant j \leqslant k \leqslant 6$, it must be true that $a_{1}+a_{j} \leqslant a_{k}$, which means that the three sides of this hexagon cannot form a triangle. Now, let $a_{1}=a_{2}=1, a_{3}=2$, $a_{4}=3, a_{5}=5, a_{6}=8$, then $a_{1}, a_{2}, \cdots, a_{8}$ satisfy all conditions.

Therefore, such a hexagon exists. Furthermore, due to the instability of an $n$-sided polygon ($n \geqslant 4$), we know that there are infinitely many such hexagons.