1. $\frac{2^{2}+1}{2^{2}-1}+\frac{3^{2}+1}{3^{2}-1}+\cdots+\frac{2005^{2}+1}{2005^{2}-1}=$
$$
=, 1.2005 \frac{1003502}{2011015} .
$$

First, consider the general term of the sum
$$
\begin{array}{l}
a_{n}=\frac{n^{2}+1}{n^{2}-1}=1+\frac{2}{(n-1)(n+1)} \\
=1+\frac{(n+1)-(n-1)}{(n-1)(n+1)}=1+\frac{1}{n-1}-\frac{1}{n+1}
\end{array}
$$

Taking $n=2,3, \cdots, k$, and summing up, we have
$$
\begin{array}{l}
\frac{2^{2}+1}{2^{2}-1}+\frac{3^{2}+1}{3^{2}-1}+\cdots+\frac{k^{2}+1}{k^{2}-1} \\
=(k-1)+1+\frac{1}{2}-\frac{1}{k}-\frac{1}{k+1} \\
=k+\frac{1}{2}-\frac{1}{k}-\frac{1}{k+1} .
\end{array}
$$

Taking $k=2005$, we get
$$
\begin{array}{l}
\text { Original expression }=2005+\frac{1}{2}-\frac{1}{2005}-\frac{1}{2006} \\
=2005 \frac{1003502}{2011015} .
\end{array}
$$