10. (20 points) Find all possible positive integers $n \geqslant 3$, such that there exist distinct positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$, for which all the products $a_{i} a_{j}(1 \leqslant i<j \leqslant n)$ can be rearranged in increasing order to form a geometric progression.
10. Notice that, when $n=3$, we can choose
$$
a_{1}=1, a_{2}=2, a_{3}=4 .
$$

Then $a_{1} a_{2}=2, a_{1} a_{3}=4, a_{2} a_{3}=8$ form a geometric sequence.
When $n=4$, we can choose
$$
a_{1}=1, a_{2}=4, a_{3}=8, a_{4}=16 \text {. }
$$

Then $a_{1} a_{2}=4, a_{1} a_{3}=8, a_{1} a_{4}=16, a_{2} a_{3}=32$, $a_{2} a_{4}=64, a_{3} a_{4}=128$ form a geometric sequence.
When $n \geqslant 5$, assume without loss of generality that $a_{1}1$. Since the first two terms of the geometric sequence $S$ are $a_{1} a_{2}$ and $a_{1} a_{3}$, and the last two terms are $a_{n-2} a_{n}$ and $a_{n-1} a_{n}$, we have
$$
\frac{a_{3}}{a_{2}}=\frac{a_{1} a_{3}}{a_{1} a_{2}}=q=\frac{a_{n-1} a_{n}}{a_{n-2} a_{n}}=\frac{a_{n-1}}{a_{n-2}} .
$$

If $n \geqslant 6$, then $a_{3} a_{n-2}=a_{2} a_{n-1}$, which contradicts the fact that $a_{3} a_{n-2}$ and $a_{2} a_{n-1}$ are different terms in the geometric sequence $S$.
If $n=5$, then $\frac{a_{3}}{a_{2}}=q=\frac{a_{4}}{a_{3}}$.
From $\frac{a_{1} a_{4}}{a_{1} a_{3}}=q$ and $\frac{a_{3} a_{5}}{a_{2} a_{5}}=q$, we know that the first three terms of the geometric sequence $S$ are $a_{1} a_{2}$, $a_{1} a_{3}$, and $a_{1} a_{4}$, and the last three terms are $a_{2} a_{5}$, $a_{3} a_{5}$, and $a_{4} a_{5}$.

Also, from $\frac{a_{2} a_{4}}{a_{2} a_{3}}=q$ and $\frac{a_{3} a_{4}}{a_{2} a_{4}}=q$, we know that $a_{2} a_{3}$, $a_{2} a_{4}$, and $a_{3} a_{4}$ are consecutive terms in the geometric sequence $S$.
If $a_{1} a_{5}$ is the fourth term of the geometric sequence $S$, then
$$
\frac{a_{5}}{a_{4}}=\frac{a_{1} a_{5}}{a_{1} a_{4}}=q=\frac{a_{3}}{a_{2}},
$$

Thus, $a_{2} a_{5}=a_{3} a_{4}$, which contradicts the fact that $a_{2} a_{5}$ and $a_{3} a_{4}$ are different terms in the geometric sequence $S$;
If $a_{1} a_{5}$ is the seventh term of the geometric sequence $S$, then
$$
\frac{a_{2}}{a_{1}}=\frac{a_{2} a_{5}}{a_{1} a_{5}}=q=\frac{a_{4}}{a_{3}},
$$

Thus, $a_{2} a_{3}=a_{1} a_{4}$, which contradicts the fact that $a_{2} a_{3}$ and $a_{1} a_{4}$ are different terms in the geometric sequence $S$.