【Analysis】Answer: 9.
Substitute $a+b$ into the equation to get: $(a+b)^{2}+a(a+b)+b=0$, rearranging gives:
$$
b^{2}+(3 a+1) b+2 a^{2}=0,
$$

Obviously, at least one of $a$ and $b$ is negative. To find the maximum value of $b$, then $a<0$. Considering $b$ as the main variable, we get:
$$
b=\frac{-(3 a+1)-\sqrt{(3 a+1)^{2}-8 a^{2}}}{2}=\frac{-(3 a+1)-\sqrt{a^{2}+6 a+1}}{2},
$$

where:
$$
a \geq 2 \sqrt{2}-3 \text { or } a \leq-(2 \sqrt{2}+3),
$$

Since $b \in \mathbb{Z}$, thus: $a^{2}+6 a+1=m^{2}, m \in \mathbb{Z}$,
which means:
$$
a^{2}+6 a+1-m^{2}=0
$$

has integer solutions.
$$
\Delta=36-4\left(1-m^{2}\right)=4\left(m^{2}+8\right)
$$

is a perfect square, let
$$
m^{2}+8=n^{2} \text {, where: } n \in \mathbb{Z} \text {, }
$$

so:
$$
(n+m)(n-m)=8=2 \times 4=(-2) \times(-4),
$$

solving gives:
$$
\left\{\begin{array}{l}
n= \pm 3, \\
m= \pm 1, \\
a=0 \text { or }-6, \\
b=-1 \text { or } 9,
\end{array}\right.
$$

Thus $b_{\text {max }}=9$, at this time $a=-6$.
8. Given $a, b \in Z$, and $a+b$ is a root of the equation $x^{2}+a x+b=0$, then the maximum possible value of $b$ is