[Solution] Let the common ratio be $r (r > 1)$ for a geometric sequence satisfying
$$
100 \leqslant aq > 1) .(p, q)=1$.
Since $\quad a r^{n-1}=\frac{a p^{n-1}}{q^{n-1}}$

and $p$ and $q$ are coprime, then $q^{n-1}$ can divide $a, \quad \frac{a}{q^{n-1}} \geqslant 1$.
To maximize the number of terms in the geometric sequence, we take $p=q+1$, thus we have
$$
100 \leqslant a < a\left(\frac{q+1}{q}\right) < \cdots < a\left(\frac{q+1}{q}\right)^{n-1} \leqslant 1000 .
$$

If $q \geqslant 3$, then
$$
1000 \geqslant a\left(\frac{q+1}{q}\right)^{n-1} \geqslant (q+1)^{n-1} \geqslant 4^{n-1} .
$$

So
$$
\begin{array}{c}
n-1 \leqslant \frac{3}{\lg 4} \\
n \leqslant 5
\end{array}
$$

If $q=1$, then
$$
\begin{array}{c}
1000 \geqslant a\left(\frac{q+1}{q}\right)^{n-1}=a \cdot 2^{n-1} \geqslant 100 \cdot 2^{n-1} . \\
n-1 \leqslant \frac{1}{\lg 2} \\
n \leqslant 4 .
\end{array}
$$

If $q=2$, then
$$
\begin{array}{c}
1000 \geqslant a\left(\frac{q+1}{q}\right)^{n-1}=a\left(\frac{3}{2}\right)^{n-1} \geqslant 100 \cdot\left(\frac{3}{2}\right)^{n-1} . \\
n-1 \leqslant \frac{1}{\lg \frac{3}{2}} \\
n \leqslant 6 .
\end{array}
$$

Therefore, the sequence can have at most 6 terms.
When $n=6$, take $q=2, r=\frac{3}{2}$, from
$$
a \cdot\left(\frac{3}{2}\right)^{5} \leqslant 1000
$$

we get
$$
100 \leqslant a \leqslant \frac{32000}{243}
$$

Take an integer $a$ that satisfies inequality (1) and $2^{5} \mid a$, we get
$$
a=128 \text {. }
$$

Thus, the sequence that satisfies the conditions is
$$
128, 192, 288, 432, 648, 972 \text {. }
$$
6.24 In a geometric sequence with a common ratio greater than 1, what is the maximum number of terms that are integers between 100 and 1000.
(4th Canadian Mathematics Competition, 1972)