10. (This question is worth 20 points) Let $a_{1}, a_{2}, a_{3}, a_{4}$ be four rational numbers, such that: $\left\{a_{i} a_{j} \mid 1 \leq i<j \leq 4\right\}=\left\{-24,-2,-\frac{3}{2},-\frac{1}{8}, 1,3\right\}$, find the value of $a_{1}+a_{2}+a_{3}+a_{4}$.
Solution: From the conditions, we know that $a_{i} a_{j} (1 \leq i<j \leq 4)$ are 6 distinct numbers, and no two of them are opposites. Therefore, the absolute values of $a_{1}, a_{2}, a_{3}, a_{4}$ are all distinct. Without loss of generality, assume $\left|a_{1}\right|<\left|a_{2}\right|<\left|a_{3}\right|<\left|a_{4}\right|$. Then, the smallest and second smallest numbers among $\left|a_{i} \| a_{j}\right| (1 \leq i<j \leq 4)$ are $\left|a_{1}\right|\left|a_{2}\right|$ and $\left|a_{1}\right|\left|a_{3}\right|$, respectively, and the largest and second largest numbers are $\left|a_{3}\right|\left|a_{4}\right|$ and $\left|a_{2}\right|\left|a_{4}\right|$, respectively. Thus, we must have
$$
\left\{\begin{array}{l}
a_{1} a_{2}=-\frac{1}{8}, \\
a_{1} a_{3}=1, \quad 10 \text { points } \\
a_{2} a_{4}=3, \\
a_{3} a_{4}=-24,
\end{array}\right.
$$

Therefore, $a_{2}=-\frac{1}{8 a_{1}}, a_{3}=\frac{1}{a_{1}}, a_{4}=\frac{3}{a_{2}}=-24 a_{1}$.
Hence, $\left\{a_{2} a_{3}, a_{1} a_{4}\right\}=\left\{-\frac{1}{8 a_{1}^{2}},-24 a_{1}^{2}\right\}=\left\{-2,-\frac{3}{2}\right\}, 15$ points. Combining $a_{1} \in \mathbb{Q}$, the only possible values for $a_{1}$ are $\pm \frac{1}{4}$.
Thus, we have $a_{1}=\frac{1}{4}, a_{2}=-\frac{1}{2}, a_{3}=4, a_{4}=-6$ or $a_{1}=-\frac{1}{4}, a_{2}=\frac{1}{2}, a_{3}=-4, a_{4}=6$.
Verification shows that both sets of solutions satisfy the conditions of the problem.
Therefore, $a_{1}+a_{2}+a_{3}+a_{4}= \pm \frac{9}{4}. 20$ points.