1. Given $m=\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\cdots+$ $\frac{1}{\sqrt{2003}+\sqrt{2005}}$. Then $50[2 m]-142=$ $\qquad$ .
(Where $[x]$ denotes the greatest integer not exceeding $x$)
2. 2008 .
From the known equation, we get
$$
m=\frac{\sqrt{2005}-1}{2} \text {. }
$$

Thus, $[2 m]=43$.
Therefore, $50[2 m]-142=2008$.