9. Let the line $l: y=k x+m$ (where $k, m$ are integers) intersect the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ at two distinct points $A, B$, and intersect the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ at two distinct points $C, D$. Does there exist a line $l$ such that the vector $\overrightarrow{A C}+\overrightarrow{B D}=0$? If it exists, how many such lines are there? If not, please explain the reason.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and solve the system $\left\{\begin{array}{l}3 x^{2}+4 y^{2}=48, \\ y=k x+m\end{array} \Rightarrow 3 x^{2}+4(k x+m)^{2}=48\right.$ $\Rightarrow\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-48=0 \Rightarrow x_{1}+x_{2}=-\frac{8 k m}{3+4 k^{2}}$.
Similarly, let $C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$, and solve the system $\left\{\begin{array}{l}3 x^{2}-y^{2}=12, \\ y=k x+m\end{array} \Rightarrow 3 x^{2}-(k x+m)^{2}=\right.$ $\Rightarrow\left(3-k^{2}\right) x^{2}-2 k m x-m^{2}-12=0 \Rightarrow x_{3}+x_{4}=\frac{2 k m}{3-k^{2}}$.
According to the problem, $-\frac{8 k m}{3+4 k^{2}}=\frac{2 k m}{3-k^{2}}$. If $k m \neq 0 \Rightarrow 3+4 k^{2}=k^{2}-3$ has no solution; if $k m=0$, since $\left\{\begin{array}{l}\Delta_{1}>0 \Rightarrow 16 k^{2}+12-m^{2}>0, \\ \Delta_{2}>0 \Rightarrow m^{2}+12-4 k^{2}>0,\end{array}\right.$ then when $k=0$, $m=0, \pm 1, \pm 2, \pm 3 ;$ when $m=0$, $m=0, \pm 1$. Therefore, there are 9 lines that satisfy the conditions.