Given that $a, \quad b, \quad c$ and $d$ are negative numbers, and $\left|x_{1}-a\right|+\left|x_{2}+b\right|+\left|x_{3}-c\right|+\left|x_{4}+d\right|=0$, then the value of $\frac{x_{1} x_{2}}{x_{3} x_{4}}$ is $\quad(\quad)$.
(A) a negative number
(B) a non-negative number
(C) a positive number
(D) a non-positive number

Translation: Given that $a, b, c, d$ are all negative numbers, and $\left|x_{1}-a\right|+\left|x_{2}+b\right|+\left|x_{3}-c\right|+\left|x_{4}+d\right|=0$, then $\frac{x_{1} x_{2}}{x_{3} x_{4}}$ is ( ).
(A) a negative number
(B) a non-negative number
(C) a positive number
(D) a non-positive number
Answer: $\mathrm{C}$.
Solution: By the definition of absolute value, we know
$$
\left|x_{1}-a\right| \geq 0,\left|x_{2}+b\right| \geq 0,\left|x_{3}-c\right| \geq 0,\left|x_{4}+d\right| \geq 0,
$$

Also, since
$$
\left|x_{1}-a\right|+\left|x_{2}+b\right|+\left|x_{3}-c\right|+\left|x_{4}+d\right|=0,
$$

Therefore,
$$
\left|x_{1}-a\right|=0,\left|x_{2}+b\right|=0,\left|x_{3}-c\right|=0,\left|x_{4}+d\right|=0 \text {. }
$$

That is
$$
x_{1}=a, x_{2}=-b, x_{3}=c, x_{4}=-d
$$

Since
$a, b, c, d$ are all negative numbers,

Thus $x_{1}, x_{3}$ are negative, and $x_{2}, x_{4}$ are positive,

Hence $\frac{x_{1} x_{2}}{x_{3} x_{4}}$ is positive.

Therefore, the answer is C.