6. The sum of $n$ consecutive natural numbers starting from the positive integer $m$ is 2004, and $(m, n)>1$ (not coprime). Then the greatest common divisor $(m, n, 2004)=$


6. 12 .

According to the problem, $(m, n)>1$ and $m+(m+1)+\cdots+(m+n-1)=2004$, that is,
$$
\frac{(2 m+n-1) n}{2}=2004 \text {. }
$$

And $(2 m+n-1) n$
$$
=1 \times 4008=3 \times 1336=167 \times 24=501 \times 8 \text {, }
$$

where $2 m+n-1$ and $n$ are one odd and one even, and $2 m+n-1$ is greater than $n$. Therefore, we have
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ 2 m + n - 1 = 4 0 0 8 , } \\
{ n = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=2004, \\
n=1 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ 2 m + n - 1 = 1 3 3 6 , } \\
{ n = 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=667, \\
n=3 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ 2 m + n - 1 = 1 6 7 , } \\
{ n = 2 4 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=72, \\
n=24 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ 2 m + n - 1 = 5 0 1 , } \\
{ n = 8 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=247, \\
n=8 .
\end{array}\right.\right.
\end{array}
$$

Given that $m$ and $n$ are not coprime, we can only take
$$
m=72=2^{3} \times 3^{2}, n=24=2^{3} \times 3 \text {. }
$$

Also, $2004=2^{2} \times 3 \times 167$. Therefore, $(m, n, 2004)=12$.