6. In $\triangle A B C$, angles $A, B, C$ satisfy $1007 A^{2}+1009 B^{2}=2016 C^{2}$, then $\triangle A B C$ is
A. Acute triangle
B. Right triangle
C. Obtuse triangle
D. Any of the above is possible
Solve:
$\frac{2016 C^{2}}{1007 A^{2}}-\frac{1009 B^{2}}{1007 A^{2}}=1$, Let $x=\frac{C}{A}, y=\frac{B}{A}$,
then $\frac{x^{2}}{\frac{1007}{2016}}-\frac{y^{2}}{\frac{1007}{1009}}=1$. As shown in the figure, the hyperbola intersects with the line
$x+y=1$, so $B+C>A$,
$B+C=A, B+C<A$ are all possible.
And $A+B+C=\pi$, meaning $\triangle ABC$ can be an acute, 
right, or obtuse triangle, so the answer is $D$.