1. Among the positive integers $1,2, \cdots, 2008$, the number of integers that can be expressed in the form $\frac{m n-1}{m-n}\left(m, n \in \mathbf{N}_{+}\right)$ is ( ).
(A) 2008
(B) 2006
(C) 2007
(D) 2004
- 1.C.

If $\frac{m n-1}{m-n}=1$, rearranging gives $(m+1)(n-1)=0$.
Solving, we get $m=-1$ or $n=1$.
Therefore, when $n=1, m>1$, $\frac{m n-1}{m-n}=\frac{m-1}{m-1}=1$.
If $\frac{m n-1}{m-n}=2$, rearranging gives $(m+2)(n-2)=-3=3 \times(-1)$.
Solving, we get $m=1, n=1$.
But $m \neq n$, thus, $\frac{m n-1}{m-n} \neq 2$.
Therefore, 2 cannot be expressed as a number in the form $\frac{m n-1}{m-n}$. If $\frac{m n-1}{m-n}=a(a \in \mathbf{Z}, a>2)$, then $m n-a m+a n-1=0$. Hence, $(m+a)(n-a)=1-a^{2}=\left(a^{2}-1\right) \times(-1)$.
Therefore, $\left\{\begin{array}{l}m+a=a^{2}-1, \\ n-a=-1,\end{array}\right.$ which is
$$
\left\{\begin{array}{l}
m=a^{2}-a-1 \\
n=a-1
\end{array}\right.
$$

Clearly, $m \neq n$.
When $m=a^{2}-a-1, n=a-1$, $\frac{m n-1}{m-n}=\frac{\left(a^{2}-a-1\right)(a-1)-1}{\left(a^{2}-a-1\right)-(a-1)}=a$.