6. Let $A=\{1,2, \cdots, 2002\}, M=\{1001,2003,3005\}$. For any non-empty subset $B$ of $A$, if the sum of any two numbers in $B$ does not belong to $M$, then $B$ is called an $M$-free set. If $A=A_{1} \cup A_{2}, A_{1} \cap A_{2}=\varnothing$, and both $A_{1}$ and $A_{2}$ are $M$-free sets, then the ordered pair $\left(A_{1}, A_{2}\right)$ is called an $M$-partition of $A$. Find the number of all $M$-partitions of $A$.
6. The answer is $2^{501}$. For $m, n \in A$, if $m+n=1001$ or 2003 or 3005, then $m$ and $n$ are said to be "related".

It is easy to see that the numbers related to 1 are only 1000 and 2002, the numbers related to 1001 and 2002 are 1 and 1003, and the numbers related to 1003 are 1000 and 2002.
Therefore, for $1,1003,1000,2002$, they must be divided into two groups $\{1,1003\},\{1000,2002\}$,
Similarly, other groups can be divided as $\{2,1004\},\{999,2001\},\{3,1005\},\{998,2000\}, \cdots,\{500, 1502\},\{501,1503\},\{1001,1002\}$.
Thus, the 2002 numbers in $A$ are divided into 501 pairs, totaling 1002 groups.

Since any number is related to the corresponding number in another group, if one group in a pair is in $A_{1}$, the other group must be in $A_{2}$. Conversely, if one group is in $A_{2}$, the other group must be in $A_{1}$, and there are no related numbers in $A_{1}$ and $A_{2}$. Therefore, the number of $M$-partitions of $A$ is $2^{501}$.