10. (20 points) Let $a_{1}, a_{2}, a_{3}, a_{4}$ be four 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 $a_{1}+a_{2}+a_{3}+a_{4}$.
$$
\begin{array}{l}
\text { From the conditions, we know that } a_{i} a_{j}(1 \leqslant i<j \leqslant 4) \text { are six distinct numbers, and no two are opposite numbers. } \\
\text { Therefore, the absolute values of } a_{1}, a_{2}, a_{3}, a_{4} \text { are all distinct. Without loss of generality, assume } \left|a_{1}\right|<\left|a_{2}\right|<\left|a_{3}\right|<\left|a_{4}\right|. \\
\text { Then, the smallest and second smallest numbers among } \left|a_{i}\right|\left|a_{j}\right|(1 \leqslant i<j \leqslant 4) \text { are } \left|a_{1}\right|\left|a_{2}\right| \text { and } \left|a_{1}\right|\left|a_{3}\right|, \\
\text { and the largest and second largest numbers are } \left|a_{3}\right|\left|a_{4}\right| \text { and } \left|a_{2}\right|\left|a_{4}\right|. \\
\text { Hence } \left\{\begin{array} { l } 
{ a _ { 1 } a _ { 2 } = - \frac { 1 } { 8 } , } \\
{ a _ { 1 } a _ { 3 } = 1 , } \\
{ a _ { 2 } a _ { 4 } = 3 , } \\
{ a _ { 3 } a _ { 4 } = - 2 4 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a_{2}=-\frac{1}{8 a_{1}}, \\
a_{3}=\frac{1}{a_{1}}, \\
a_{4}=\frac{3}{a_{2}}=-24 a_{1}
\end{array}\right.\right. \\
\Rightarrow\left\{a_{2} a_{3}, a_{1} a_{4}\right\}=\left\{-2,-\frac{3}{2}\right\} \text {. } \\
\end{array}
$$

Combining with \( a_{1} \in \mathbf{Q} \), it is only possible that \( a_{1} = \pm \frac{1}{4} \).
From this, it is easy to see that
$$
\begin{array}{l}
\left(a_{1}, a_{2}, a_{3}, a_{4}\right) \\
=\left(\frac{1}{4},-\frac{1}{2}, 4,-6\right) \text { or }\left(-\frac{1}{4}, \frac{1}{2},-4,6\right) .
\end{array}
$$

Upon verification, both solutions satisfy the conditions.
Thus, \( a_{1} + a_{2} + a_{3} + a_{4} = \pm \frac{9}{4} \).
$$