红眼修士

设实际有 $k$ 位红眼修士。

游客的公开声明成为“公共知识”，触发归纳推理：

• 若 $k=1$，那位红眼修士看不到别人红眼，听到声明后立刻知道自己红眼，于是第 1 天午夜自尽。
• 假设当红眼人数为 $k-1$ 时，会在第 $k-1$ 天午夜同时自尽。
• 当红眼人数为 $k$ 时，每位红眼修士都能看到 $k-1$ 位红眼。如果自己不是红眼，则红眼人数应为 $k-1$，按归纳假设应在第 $k-1$ 天午夜发生自尽；但他观察到前 $k-1$ 天都无人自尽，于是第 $k$ 天得知自己也红眼，并在第 $k$ 天午夜自尽。

因此 $\boxed{\text{若共有 }k\text{ 位红眼，则第 }k\text{ 天午夜这 }k\text{ 人同时自尽。}}$

在常见版本里 $k=100$，所以第 $\boxed{100}$ 天午夜 100 位红眼修士同时自尽。

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英文解析

Suppose there are actually $k$ red-eyed monks.

The tourist's public declaration becomes "common knowledge," triggering inductive reasoning:

• If $k=1$, that red-eyed monk sees no other red-eyed monks and, upon hearing the declaration, immediately knows he is red-eyed, thus committing suicide at midnight on Day 1.
• Assume that when the number of red-eyed monks is $k-1$, they will all commit suicide simultaneously at midnight on Day $k-1$.
• When the number of red-eyed monks is $k$, each red-eyed monk sees $k-1$red-eyed monks. If he were not red-eyed, the number of red-eyed monks would be $k-1$, and according to the inductive hypothesis, they would commit suicide at midnight on Day $k-1$. However, he observes that no one commits suicide during the first $k-1$days, so he deduces on Day $k$that he is also red-eyed and commits suicide at midnight on Day $k$.

Therefore $\boxed{\text{If there are } k \text{ red-eyed monks, then these } k \text{ people commit suicide simultaneously at midnight on Day } k.}$

In the common version, $k=100$, so 100 red-eyed monks commit suicide simultaneously at midnight on Day $\boxed{100}$.