1. Given $x\left(y-\frac{1}{x}\right)=2008$, where $x$ and $y$ are positive integers. Then the sum of the maximum and minimum values of $x+y$ is ( ).
(A) 2010
(B) 2100
(C) 2008
(D) 2000
-.1.B.
From the given, we have
$$
\begin{array}{l}
2008=x\left(y-\frac{1}{x}\right) \\
\Rightarrow x y=2009=1 \times 2009=7 \times 7 \times 41 .
\end{array}
$$

Thus, $(x+y)_{\text {max }}=1+2009=2010$,
$$
(x+y)_{\text {min }}=49+41=90 \text {. }
$$

Therefore, the sum of the maximum and minimum values of $x+y$ is 2100.