Three. (20 points) Place the 2004 positive integers 1, 2, $\cdots$, 2004 randomly on a circle. By counting the parity of all adjacent 3 numbers, it is found that there are 600 groups where all 3 numbers are odd, and 500 groups where exactly 2 numbers are odd. How many groups have exactly 1 odd number? How many groups have no odd numbers at all?
Three, let there be $x$ groups with exactly 1 odd number. Then, all are not odd in $2004-600-500-x=904-x$.

Starting from a certain number on the circle, sequentially denote them as $x_{1}, x_{2}$, $\begin{array}{l} \cdots, x_{2000} \text {, let } \\ y_{i}=\left\{\begin{array}{cc}-1, & \text { when } x_{i} \text { is odd, } \\ 1, & \text { when } x_{i} \text { is even. }\end{array}\right.\end{array}$
Then $y_{1}+y_{2}+\cdots+y_{2004}=0$, and
$A_{i}=y_{i}+y_{i+1}+y_{i+2}$
$=\left\{\begin{array}{cl}-3, & \text { when } x_{i} 、 x_{i+1} 、 x_{i+2} \text { are all odd, } \\ -1, & \text { when } x_{i} 、 x_{i+1} 、 x_{i+2} \text { have exactly 2 odd numbers, } \\ 1, & \text { when } x_{i} 、 x_{i+1} 、 x_{i+2} \text { have exactly 1 odd number, } \\ 3, & \text { when } x_{i} 、 x_{i+1} 、 x_{i+2} \text { are all even. }\end{array}\right.$
$\left(x_{2000+i}=x_{i}\right.$. $)$
$$
\begin{array}{l}
\text { Then } 0=3\left(y_{1}+y_{2}+\cdots+y_{2000}\right) \\
=A_{1}+A_{2}+\cdots+A_{2001} \\
=-3 \times 600-500+x+3(904-x) .
\end{array}
$$

Then $0=3\left(y_{1}+y_{2}+\cdots+y_{2001}\right)$

Solving, we get $x=206,904-x=698$.
Therefore, there are 206 groups with exactly 1 odd number, and 698 groups with none being odd.