2. If $(m+6)(m+5)-(n+6)(n+5)=$ $2006 m(m-1)-2006 n(n-1)$, and $m \neq n$, then among the following four conclusions, the correct one is ( ).
(A) $m+n$ is not necessarily a constant
(B) $m+n=2006$
(C) $m+n=\frac{2017}{4010}$
(D) $m+n=\frac{2017}{2005}$
2. D.

From the problem, we have
$$
\begin{array}{l}
(m+6)(m+5)-2006 m(m-1) \\
=(n+6)(n+5)-2006 n(n-1) . \\
\text { Let } f(x)=(x+6)(x+5)-2006 x(x-1) .
\end{array}
$$

Then $f(x)$ is a quadratic function, and
$$
f(m)=f(n)(m \neq n) \text {. }
$$

Therefore, the points $(m, f(m))$ and $(n, f(n))$ must be symmetric about the axis of symmetry $x=\frac{2017}{4010}$ of the quadratic function $y=f(x)$.
Thus, $\frac{m+n}{2}=\frac{2017}{4010}$, which means $m+n=\frac{2017}{2005}$.