Answer: 26.
When $n=2 k+1$, the number of trapezoids is $(2 k+1) C_{k}^{2}$, from which we know the maximum value of $k$ is 12.
Therefore, the maximum value of $n$ is 25.
When $n=4 k$, removing the rectangles yields the number of trapezoids as $2 k\left(C_{2 k-1}^{2}+C_{2 k}^{2}\right)-C_{2 k}^{2}$, from which we find the maximum value of $k$ is 6, making the maximum value of $n$ 24.

When $n=4 k+2$, the number of trapezoids is $(2 k+1)\left(C_{2 k}^{2}+C_{2 k+1}^{2}\right)-C_{2 k+1}^{2}$, from which we find the maximum value of $k$ is 6.
Thus, the maximum value of $n$ is 26.
In summary, the maximum value of $n$ is 26.
8、On the circumference, there are $n$ equally divided points. The number of non-congruent trapezoids formed by choosing four of these points as vertices is no more than 2012. Then the maximum value of the positive integer $n$ is $\qquad$