6. Given $a_{1}, a_{2}, \cdots, a_{2011}$ is a sequence of distinct positive integers. If the order of these 2011 numbers is arbitrarily changed and denoted as $b_{1}, b_{2}, \cdots, b_{2011}$. Then the value of the number
$$
M=\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right) \cdots\left(a_{2011}-b_{2011}\right)
$$
must be $(\quad)$.
(A) 0
(B) 1
(C) odd number
(D) even number
6. D.

Assume $M$ is odd. Then
$$
a_{i}-b_{i}(i=1,2, \cdots, 2011)
$$

must all be odd, and their sum is also odd;
$$
\begin{array}{l} 
\text { but }\left(a_{1}-b_{1}\right)+\left(a_{2}-b_{2}\right)+\cdots+\left(a_{2011}-b_{2011}\right) \\
=\left(a_{1}+a_{2}+\cdots+a_{2011}\right)- \\
\left(b_{1}+b_{2}+\cdots+b_{2011}\right) \\
=
\end{array}
$$

is even, a contradiction.
Therefore, $M$ is even.