6. If the perimeter of a hexagon is 20, all sides are integers, 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.
6. (D)
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) $a_{1} \leqslant a_{2}, a_{1}+a_{2} \leqslant a_{3}, a_{2}+a_{3} \leqslant a_{4}$; $a_{3}+a_{4} \leqslant a_{5}, a_{4}+a_{5} \leqslant a_{6}$;
(3) $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}>a_{6}$.
Then a hexagon with side lengths $a_{1}, a_{2}, \cdots, a_{6}$ will meet the requirements.
In fact, for any three integers $1 \leqslant i<j<k \leqslant 6$, there must be $a_{i}+a_{j} \leqslant a_{k}$, indicating that no three sides of this hexagon 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 conditions.

Therefore, such a hexagon exists at least one. Furthermore, due to the instability of an $n$-sided polygon $(n \geqslant 4)$, it is known that there are infinitely many such hexagons.