9.01 Try to find how many ways there are to divide the set $\left\{2^{0}, 2^{1}, 2^{2}, \cdots, 2^{2005}\right\}$ into two non-empty disjoint subsets $A$ and $B$, such that the equation $x^{2}-S(A) x+S(B)=0$ has integer roots, where $S(M)$ denotes the sum of all elements in the set $M$?
There can be 1003 solutions.
Let $x_{1} \leqslant x_{2}$ be the roots of the equation, then $x_{1}, x_{2} \in \mathbf{Z}$, and $x_{1}+x_{2}=S(A)$, $x_{1} x_{2}=S(B)$. Therefore, $\left(x_{1}+1\right)\left(x_{2}+1\right)=S(B)+S(A)+1=1+2$ $+4+\cdots+2^{2005}+1=2^{2006}$. This implies $x_{1}+1=2^{k}, x_{2}+1=$ $2^{2006-k}$, where $k=1,2, \cdots, 2003$.

Conversely, when $x_{1}+1=2^{k}, x_{2}+1=2^{2006-k}$, they are the roots of the equation $x^{2}-$ $p x+q$, where $p=2^{k}+2^{2006-k}-2, q=2^{2006}-1-p$. The number $p$ has a unique binary representation, in which the highest power of 2 does not exceed 2005; and since $p+q=2^{2006}-1$, the places where $p$ has a 1 in its binary representation, $q$ has a 0, and vice versa. Thus, for each $k(1 \leqslant k \leqslant 1003)$, there exists a unique partition $(A, B)$ such that the roots of the equation are exactly $x_{1}$ and $x_{2}$.