6. Given the set
$$
A=\left\{x \mid x=a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}\right\} \text {, }
$$

where, $a_{i} \in\{0,1, \cdots, 6\}(i=0,1,2,3)$, and $a_{3} \neq 0$.
If positive integers $m 、 n \in A$, and $m+n=2010(m>n)$, then the number of positive integers $m$ that satisfy the condition is $\qquad$.

6. 662 .

According to the problem, we know that $m$ and $n$ are four-digit numbers in base 7, and the largest four-digit number in base 7 is
$$
6 \times 7^{3}+6 \times 7^{2}+6 \times 7+6=2400,
$$

the smallest one is $1 \times 7^{3}=343$.
Since $m+n=2010(m>n)$, therefore,
$$
1006 \leqslant m \leqslant 1667 \text {. }
$$

Thus, the number of positive integers $m$ that meet the condition is
$$
1667-1006+1=662 \text { (numbers). }
$$