6. Given that for each integer $m$ satisfying $0<m<2001$, there exists an integer $k$ such that $\frac{m}{2001}<\frac{k}{n}<\frac{m+1}{2002}$ holds, where $n$ is a positive integer, then the minimum possible value of $n$ is
A. 2003
B. 2008
C. 3106
D. 4003
6. D First, it is easy to verify that when $\frac{m}{2001}0$, $\frac{x}{x+1}$ is strictly monotonically increasing. Noting that $n>2002$, we have $k \leqslant n-2$. However, $\frac{2000}{2001}4002$. In conclusion, $n=4003$.