51 Let the set $S=\{100,101,102, \cdots, 999,1000\}, A=\left\{a_{1}, a_{2}, a_{3}, \cdots, a_{n-1}, a_{n}\right.$ $\mid a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers, and $\left.\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\cdots=\frac{a_{n}}{a_{n-1}}=q>1\right\}$. Try to find the maximum possible number of elements in the intersection $S \cap A$.
51 Let $a_{1}, a_{n} \in S$, then $q^{n-1}=\frac{a_{n}}{a_{1}} \in \mathbf{Q}$. If $t$ is the smallest positive integer such that $q^{t} \in \mathbf{Q}$, then for any positive integer $k$ such that $q^{k} \in \mathbf{Q}$, by the division algorithm $k=c t+d$ (where $c, d$ are non-negative integers and $d<t$), thus $q^{d}=q^{k-c t} \in \mathbf{Q}$. By the minimality of $t$, $d=0$. Therefore, $t \mid k$. Hence, for any $-a_{k+1} \in S, a_{k+1}=$ $a_{1} q^{k}=a_{1}\left(q^{t}\right)^{k t}$. Thus, we can assume $q$ is a rational number (otherwise, replace $q$ with $q^{t}$).
Let $q=\frac{a}{b}$, where $a, b$ are coprime natural numbers. If $b \geqslant 3$, then
$$
a \geqslant 4, a_{n} \pm a_{1} q^{n-1}=\frac{a_{1}}{b^{n-1}} \times a^{n-1} \geqslant a^{n-1} \geqslant 4^{n-1} .
$$

From $1000 \geqslant 4^{n-1}$, we get $n \leqslant 5$. If $b=1$, then
$$
a \geqslant 2,1000 \geqslant a_{n}=a_{1} q^{n-1} \geqslant 100 \cdot a^{n-1} \geqslant 100 \cdot 2^{n-1} .
$$

Thus, $n \leqslant 4$. If $b=2$, then
$$
a \geqslant 3,1000 \geqslant a_{1} q^{n-1} \geqslant 100\left(\frac{3}{2}\right)^{n-1},
$$

Thus, $n \leqslant 6$. In summary, $|S \cap A| \leqslant 6$.
On the other hand, 128, 192, 288, 432, 648, 972 form a geometric sequence with a common ratio of $\frac{3}{2}$. Therefore, the maximum value of $|S \cap A|$ is 6.