6. In a graduation photo at a school, 100 students and teachers from two graduating classes of the ninth grade are to be arranged on steps in a trapezoidal formation (with more people in the front rows than in the back rows, and the number of rows being greater than or equal to 3). The photographer requires that the number of people in each row must be consecutive positive integers, so that each person in the back row can stand in the gap between two people in the front row. How many arrangements satisfy these requirements? 
(A) 1
(B) 2
(C) 4
(D) 0
6. B.

Let the last row have $k$ people, with a total of $n$ rows. Then the number of people in each row from back to front are $k, k+1, \cdots, k+(n-1)$.
According to the problem,
$$
k n+\frac{n(n-1)}{2}=100,
$$

which simplifies to $n[2 k+(n-1)]=200$.
Since $k$ and $n$ are both positive integers, and $n \geqslant 3$, we have $n<2 k+(n-1)$,

and $n$ and $2 k+(n-1)$ have different parities.
Decomposing 200 into prime factors, we find that $n=5$ or 8.
When $n=5$, $k=18$;
When $n=8$, $k=9$.
Therefore, there are two different solutions.