9.6. How many ways are there to divide the set
$$
\left\{2^{0}, 2^{1}, 2^{2}, \cdots, 2^{2000}\right\}
$$

into two non-empty disjoint subsets $A$ and $B$, such that the equation $x^{2}-S(A) x+S(B)=0$ has integer roots? Here, $S(M)$ denotes the sum of all elements in the set $M$.
9.6.1 003 types.

Let $x_{1} \leqslant x_{2}$ be the roots of the equation, then
$$
x_{1}+x_{2}=S(A), x_{1} x_{2}=S(B),
$$

and $x_{1}, x_{2} \in \mathbf{Z}_{+}$. Therefore,
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2}+1\right)=S(B)+S(A)+1 \\
=1+2+4+\cdots+2^{200}+1=2^{2006},
\end{array}
$$

which means $x_{1}+1=2^{k}, x_{2}+1=2^{2000-k}, k=1,2, \cdots, 1003$.
Conversely, when $x_{1}+1=2^{k}, x_{2}+1=2^{2000-k}$, they are the roots of the equation $x^{2}-p x+q=0$, where $p=2^{k}+2^{2000-k}-2$, $q=2^{2006}-1-p$. The number $p$ has a unique binary representation, in which the highest power of 2 does not exceed 2005. Since $p+q=2^{2006}-1$, in the binary representation of $p$, where there is a 1, there is a 0 in the binary representation of $q$, and vice versa. Therefore, for each $k(1 \leqslant k \leqslant 1003)$, there exists a unique partition $(A, B)$ such that the roots of the equation are exactly $x_{1}, x_{2}$.