Example 3: Party A ranks $n(n \geqslant 3)$ players by strength but does not disclose the ranking to Party B. Party B arbitrarily selects three players to ask Party A, and then Party A can selectively tell Party B which of the three players is the weakest or the strongest. Party B can repeatedly ask questions. If Party B can find a sequence $S_{1}, S_{2}, \cdots, S_{N}$, such that for any player $S_{i}(1 \leqslant i \leqslant N)$, player $S_{i}$ is stronger than player $S_{i+1}$, find the maximum value of $N$.
【Analysis】First, conduct a mathematical experiment.
When $n=3$, there are only three players $A, B, C$, and a sequence of two players can be arranged, so $N=2$.

When $n=4$, let's assume the players asked are $A, B, C$, and assume player $A$ is the strongest. Then ask players $A, C, D$.
(1) If player $A$ is the strongest, then continue to ask players $B, C, D$, to get a sequence of three players.
(2) If player $D$ is the strongest, the sequence of three players is $D, A, C$.
(3) If player $D$ is the weakest, the sequence of three players is $A, C, D$.
(4) If player $C$ is the weakest, then players $B, C$ cannot be distinguished, and players $A, D$ cannot be distinguished, only a sequence of two players can be obtained.
So $N=2$.
Discuss $n$ by odd and even.
When $n$ is odd, let's assume $n=2k+1$. First, prove: Player B can find a sequence of $k+1$ determined players.

If there are two players $x, y$, no matter how Player B asks, they cannot know the strength of players $x, y$, then players $x, y$ are called a "problem pair". Clearly, any player $x$ can belong to at most one problem pair (for example, in $n=4$, situation (4) players $A, D$ and players $B, C$ are problem pairs). Otherwise, if players $x, y$ and players $x, z$ are both problem pairs, then Player B can choose three players $x, y \geq z$ to ask, at this time, Player A needs to say which one is the strongest or the weakest. This contradicts the fact that players $x, y$ and players $x, z$ are both problem pairs. Therefore, there are at most $k$ problem pairs.

Assume there are $m (m \leq k)$ problem pairs, and there are $2k+1-2m$ players left, at this time, $2k+1-2m$ players and $m$ problem pairs each take one player, a total of $2k+1-m$ players, there are no problem pairs. Thus, Player B can sort $2k+1-m$ players.

Since $2k+1-m \geq k+1$, Player B can sort these $k+1$ players.
Therefore, $N \geq k+1$.
Below, we prove by example: $N=k+1$.
Player A can use a strategy to make it impossible for Player B to find a sequence of $k+2$ players. Let Player A sort these $2k+1$ players from weak to strong as $c_{1}, c_{2}, \cdots, c_{2k+1}$, and set
$\left\{c_{1}, c_{2}\right\},\left\{c_{3}, c_{4}\right\}, \cdots,\left\{c_{2k-1}, c_{2k}\right\}$ as $k$ problem pairs. If Player B asks Player A about the strength of players $c_{2i-1}, c_{2i}, c_{j} (1 \leq i \leq k)$, then Player A only says that $c_{j}$ is the strongest or the weakest, i.e., Player A does not reveal the strength relationship between $c_{2i-1}, c_{2i}$. At this time, $2k+1$ players can be divided into
$\left\{c_{1}, c_{2}\right\},\left\{c_{3}, c_{4}\right\}, \cdots,\left\{c_{2k-1}, c_{2k}\right\},\left\{c_{2k+1}\right\}$ a total of $k+1$ groups. For any sequence of $k+2$ players, by the pigeonhole principle, it is known that there must be a pair of players $c_{2i-1}, c_{2i}$, Player B does not know their strength order, i.e., Player B cannot sort these $k+2$ players. Therefore,
$$
N_{\text {max }}=k+1=\left\lceil\frac{n}{2}\right\rceil \text {, }
$$

where $\lceil x\rceil$ represents the smallest integer not less than the real number $x$.
Similarly, when $n$ is even, let's assume $n=2k$. It is easy to prove that Player B can find a sequence of $k$ determined players, so there are at most $k$ problem pairs.

Assume there are $m (m \leq k)$ problem pairs, and there are $2k-2m$ players left, at this time, $2k-2m$ players and $m$ problem pairs each take one player, a total of $2k-m$ players, there are no problem pairs. Thus, Player B can sort $2k-m$ players. Since $2k-m \geq k$, Player B can sort these $k$ players.
Therefore, $N \geq k$.
Below, we prove by example: $N=k$.
If Player A uses the above strategy, then Player B cannot find a sequence of $k+1$ players. At this time, $2k$ players can be divided into
$\left\{c_{1}, c_{2}\right\},\left\{c_{3}, c_{4}\right\}, \cdots,\left\{c_{2k-1}, c_{2k}\right\}$

a total of $k$ groups. For any sequence of $k+1$ players, by the pigeonhole principle, it is known that there must be a pair of players $c_{2i-1}, c_{2i}$, Player B does not know their strength, i.e., Player B cannot sort these $k+1$ players. Therefore,
$$
N_{\max }=k=\left[\frac{n}{2}\right],
$$

where $[x]$ represents the largest integer not greater than the real number $x$.
$$
\text { In summary, } N_{\max }=\left\lceil\frac{n}{2}\right\rceil=\left[\frac{n+1}{2}\right] \text {. }
$$