$32 \cdot 69$ 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) is unique.
(C) is finite, but not unique.
(D) has infinitely many.
(China Junior High School Mathematics League, 1990)
[Solution] If we can find 6 integers $a_{1}, a_{2}, \cdots, a_{6}$, such that:
(1) $a_{1}+a_{2}+\cdots+a_{6}=20$;
(2) $a_{1} \leqslant a_{2}, \quad a_{1}+a_{2} \leqslant a_{3}, \quad a_{2}+a_{3} \leqslant a_{4}, \quad a_{3}+a_{4} \leqslant a_{5}$, $a_{4}+a_{5} \leqslant a_{6}$;
(3) $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}>a_{6}$.

Then a 6-sided polygon with side lengths $a_{1}, a_{2}, \cdots, a_{6}$ will meet the requirements.
In fact, for any three integers $1 \leqslant i<j<k \leqslant 6$, it must be true that $a_{i}+a_{j} \leqslant a_{k}$, which means that no three sides of this 6-sided polygon can form a triangle.

Now take $a_{1}=a_{2}=1, a_{3}=2, a_{4}=3, a_{5}=5, a_{6}=8$. Then $a_{1}$, $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ satisfy all the conditions.

Therefore, such a hexagon exists at least once, and by the instability of $n$-sided polygons $(n \leqslant 4)$, it is known that there are infinitely many such hexagons.
Hence, the answer is $(D)$.