1. Calculate $\frac{20042003^{2}+1}{20042002^{2}+20042004^{2}}=$
$$
\text { II. } 1 . \frac{1}{2} \text {. }
$$

Let $a=20042003$, then
$$
\text { Original expression }=\frac{a^{2}+1}{(a-1)^{2}+(a+1)^{2}}=\frac{a^{2}+1}{2\left(a^{2}+1\right)}=\frac{1}{2} \text {. }
$$