10. Let $a_{1}, a_{2}, a_{3}, a_{4}$ be 4 rational numbers such that
$$
\left\{a_{i} a_{j} \mid 1 \leqslant i<j \leqslant 4\right\}=\left\{-24,-2,-\frac{3}{2},-\frac{1}{8}, 1,3\right\},
$$

Find the value of $a_{1}+a_{2}+a_{3}+a_{4}$.
According to the problem, let's assume $a_{1}<a_{2}<0<a_{3}<a_{4} \Rightarrow a_{1} a_{4}=-24$.
Since the two positive numbers $a_{1} a_{2}, a_{3} a_{4}$ are 1 and 3, respectively, then $a_{1} a_{2} a_{3} a_{4}=3 \Rightarrow a_{2} a_{3}=-\frac{1}{8}$. When $a_{1} a_{2}=1, a_{3} a_{4}=3$, $a_{1}^{2} a_{2} a_{4}=-24 \Rightarrow a_{2} a_{4}=-\frac{3}{2}$
$\Rightarrow a_{1}=-4, a_{2}=-\frac{1}{4}, a_{3}=\frac{1}{2}, a_{4}=6 \Rightarrow a_{1}+a_{2}+a_{3}+a_{4}=\frac{9}{4}$;
When $a_{1} a_{2}=3, a_{3} a_{4}=1$, $a_{1}^{2} a_{2} a_{4}=-72 \Rightarrow a_{2} a_{4}=-2$
$\Rightarrow a_{1}=-6, a_{2}=-\frac{1}{2}, a_{3}=\frac{1}{4}, a_{4}=4 \Rightarrow a_{1}+a_{2}+a_{3}+a_{4}=-\frac{9}{4}$.
In summary, the value of $a_{1}+a_{2}+a_{3}+a_{4}$ is $\frac{9}{4}$ or $-\frac{9}{4}$.