\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{n\left(n+2\right)}\)
\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{n+2-n}{n\left(n+2\right)}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+2}\)
\(=1-\frac{1}{n+2}< \frac{2003}{2004}\)
\(\Leftrightarrow\frac{1}{n+2}>\frac{1}{2004}\)
\(\Leftrightarrow0< n+2< 2004\)