(9) (14 points) 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.
9 From $\left\{\begin{array}{l}y=k x+m, \\ \frac{x^{2}}{16}+\frac{y^{2}}{12}=1,\end{array}\right.$ eliminating $y$ and simplifying, we get
$$
\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-48=0 .
$$

Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, then $x_{1}+x_{2}=-\frac{8 k m}{3+4 k^{2}}$.
$$
\Delta_{1}=(8 k m)^{2}-4\left(3+4 k^{2}\right)\left(4 m^{2}-48\right)>0 \text {. }
$$

From $\left\{\begin{array}{l}y=k x+m, \\ \frac{x^{2}}{4}-\frac{y^{2}}{12}=1,\end{array}\right.$ eliminating $y$ and simplifying, we get
$$
\left(3-k^{2}\right) x^{2}-2 k m x-m^{2}-12=0 \text {. }
$$

Let $C\left(x_{3}, y_{3}\right)$ and $D\left(x_{4}, y_{4}\right)$, then $x_{3}+x_{4}=\frac{2 k m}{3-k^{2}}$.
$$
\Delta_{2}=(-2 k m)^{2}+4\left(3-k^{2}\right)\left(m^{2}+12\right)>0 .
$$

Since $\overrightarrow{A C}+\overrightarrow{B D}=0$, we have $\left(x_{4}-x_{2}\right)+\left(x_{3}-x_{1}\right)=0$, and at this time $\left(y_{4}-y_{2}\right)+\left(y_{3}-y_{1}\right)=0$. From $x_{1}+x_{2}=x_{3}+x_{4}$, we get
$$
-\frac{8 k m}{3+4 k^{2}}=\frac{2 k m}{3-k^{2}} \text {. }
$$

Thus, $2 k m=0$ or $-\frac{4}{3+4 k^{2}}=\frac{1}{3-k^{2}}$. Solving the above equation, we get $k=0$ or $m=0$. When $k=0$, from (1) and (2) we get $-2 \sqrt{3}<m<2 \sqrt{3}$. Since $m$ is an integer, the values of $m$ are $-3,-2,-1,0,1,2,3$. When $m=0$, from (1) and (2) we get $-\sqrt{3}<k<\sqrt{3}$. Since $k$ is an integer, the values of $k$ are $-1,0,1$. Therefore, there are 9 lines that satisfy the conditions.