6. (3 points) In the case where $\frac{810 \times 811 \times 812 \times \cdots \times 2010}{810^{n}}$ is a positive integer, the maximum value of $n$ is $\qquad$
【Analysis】This problem can be solved by decomposing 810 into prime factors, obtaining $810=2 \times 3 \times 3 \times 3 \times 3 \times 5$, and then finding the multiples of the prime factors of 810 in the numerators.

【Solution】Solution: $\because 810=2 \times 3 \times 3 \times 3 \times 3 \times 5, 811--2010$ contains a total of 1200 numbers, with 600 numbers having a factor of 2, 240 numbers having a factor of 5, 400 numbers having a factor of 3, 133 numbers having a factor of 9, 44 numbers having a factor of 27, 14 numbers having a factor of 81, 5 numbers having a factor of 243, and 1 number having a factor of 729.
The numbers containing a factor of 3 total $(400+133+44+14+5+1)=597$, $597 \div 4=149 \cdots 3$, $149+1=150$.

Therefore, the answer is: 150.