9. Given the polynomial with integer coefficients
$$
\begin{array}{l}
f(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+a_{3} x^{2}+a_{4} x+a_{5} . \\
\text { If } f(\sqrt{3}+\sqrt{2})=0, f(1)+f(3)=0, \\
f(-1)=
\end{array}
$$
9. 24 .

Let $x_{0}=\sqrt{3}+\sqrt{2}$. Then, $\left(x_{0}-\sqrt{3}\right)^{2}=2$.
$$
\begin{array}{l}
\text { Then } x_{0}^{2}-2 \sqrt{3} x_{0}+3=2 \\
\Rightarrow\left(2 \sqrt{3} x_{0}\right)^{2}=\left(x_{0}^{2}+1\right)^{2} \\
\Rightarrow x_{0}^{4}-10 x_{0}^{2}+1=0 .
\end{array}
$$

Thus, $x_{0}=\sqrt{3}+\sqrt{2}$ is a root of the polynomial
$$
g(x)=x^{4}-10 x^{2}+1
$$

Furthermore, $x_{0}=\sqrt{3}+\sqrt{2}$ cannot be a zero of any cubic or quadratic polynomial with integer coefficients, hence $g(x)$ divides $f(x)$, which means there exists an integer $r$ such that
$$
\begin{array}{l}
f(x)=g(x)(x-r) \\
=\left(x^{4}-10 x^{2}+1\right)(x-r) .
\end{array}
$$

Therefore, $f(1)=-8(1-r), f(3)=-8(3-r)$.
Given $f(1)+f(3)=0$, we get $r=2$.
Thus, $f(-1)=8(1+r)=24$.