2. Divide the positive integers $1,2, \cdots, 2008$ into 1004 groups: $a_{1}, b_{1} ; a_{2}, b_{2} ; \cdots ; a_{1004}, b_{1004}$, and satisfy
$$
a_{1}+b_{1}=a_{2}+b_{2}=\cdots=a_{1004}+b_{1004} \text {. }
$$

For all $i(i=1,2, \cdots, 1004), a_{i} b_{i}$ the maximum value is
2.1009020 .

Notice that
$$
\begin{array}{l}
a_{i} b_{i}=\frac{1}{4}\left[\left(a_{i}+b_{i}\right)^{2}-\left(a_{i}-b_{i}\right)^{2}\right], \\
a_{i}+b_{i}=\frac{(1+2008) \times 1004}{1004}=2009 .
\end{array}
$$

To maximize the value of $a_{i} b_{i}$, the value of $a_{i}-b_{i}$ must be minimized, and the minimum value of $a_{i}-b_{i}$ is 1. At this point, $a_{i}+b_{i}=2009, a_{i}-b_{i}=1$. Thus, $a_{i}=1005, b_{i}=1004$, and the maximum value of $a_{i} b_{i}$ is
$$
1005 \times 1004=1009020 \text {. }
$$