8、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 $C, D$, such that the vector $\overrightarrow{A C}+\overrightarrow{B D}=\overrightarrow{0}$. The number of lines that satisfy the above conditions is $\qquad$
8. 9

Analysis: Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$, and suppose $A, B, C, D$ are collinear, then $\overrightarrow{A C}+\overrightarrow{B D}=\overrightarrow{0}$ is equivalent to $x_{1}+x_{2}=x_{3}+x_{4}$. By combining the equation of the line $l$ with the equation of the ellipse, we get: $\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-48=0$. By Vieta's formulas, we know: $x_{1}+x_{2}=-\frac{8 k m}{3+4 k^{2}}$. Similarly, we get: $x_{3}+x_{4}=\frac{2 k m}{3-k^{2}}$. Therefore, from $x_{1}+x_{2}=x_{3}+x_{4}$, we know that $k=0$ or $m=0$ or
$$
\frac{4}{3+4 k^{2}}=\frac{1}{k^{2}-3} \text { . }
$$
(1) When $k=0$, it is easy to see that $m=0, \pm 1, \pm 2, \pm 3$ satisfy the conditions.
(2) When $m=0$, it is easy to see that $k=0, \pm 1$ satisfy the conditions.
(3) When $\frac{8}{3+4 k^{2}}=\frac{2}{k^{2}-3}$, there are no positive integer solutions.

In summary, there are 9 lines $l$ that satisfy the conditions.