iii. (Full marks 25 points) A mall installs an escalator between the first and second floors, which moves upwards at a uniform speed. A boy and a girl start walking up the escalator to the second floor at the same time (the escalator itself is also moving). If the boy and the girl are both considered to be moving at a uniform speed, and the boy walks twice as many steps per minute as the girl, it is known that the boy walked 27 steps to reach the top of the escalator, while the girl walked 18 steps to reach the top (assuming the boy and girl each step up one step at a time).
(i) How many steps are visible on the escalator?
(ii) If there is a staircase near the escalator that leads from the second floor to the first floor, with the same number of steps as the escalator, and the two children each walk down the stairs at their original speeds, then reach the bottom and ride the escalator again (ignoring the distance between the escalator and the stairs), how many steps will the boy have walked when he catches up to the girl for the first time?
(i) Let the girl's speed be $x$ levels/min, the escalator's speed be $y$ levels/min, and the stairs have $s$ levels. Then the boy's speed is $2x$ levels/min. According to the problem, we have:
$$
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y} ;
\end{array}\right.
$$

Solving, we get $\frac{3}{4}=\frac{s-27}{s-18}$.
Solving for $s$, we get $s=54$ (levels). That is, the escalator has 54 levels exposed.
(ii) Let the boy pass the escalator $m$ times and the stairs $n$ times when he first catches up with the girl, then the girl has passed the escalator $(m-1)$ times and the stairs $(n-1)$ times.

From the system of equations in (i), we can solve for $y=2x$, so the boy ascends the escalator at $4x$ levels/min, and the girl ascends the escalator at $3x$ levels/min.
$$
\frac{54 m}{4 x}+\frac{54 n}{2 x}=\frac{54(m-1)}{3 x}+\frac{54(n-1)}{x},
$$
$\frac{m}{4}+\frac{n}{2}=\frac{m-1}{3}+\frac{n-1}{1}$, which simplifies to $6 n+m=16$.
And $m, n$ must be positive integers, and $0 \leqslant m-n \leqslant$
1.
\begin{tabular}{c|ccccc}
$m$ & 1 & 2 & 3 & 4 & 5 \\
\hline$n=\frac{16-m}{6}$ & $2 \frac{1}{2}$ & $2 \frac{1}{3}$ & $2 \frac{1}{6}$ & 2 & $1 \frac{5}{6}$
\end{tabular}
\begin{tabular}{c|cc}
$n$ & 1 & 2 \\
\hline$m=16-6 n$ & 10 & 4
\end{tabular}

Clearly, it can only be $m=3, n=2 \frac{1}{6}$.
Thus, the number of levels the boy has walked
$$
=3 \times 27+2 \frac{1}{6} \times 54=198 \text { (levels). }
$$