老虎围羚羊

这与“狮子与肉”的经典归纳题完全同型。

若某只老虎先吃掉羚羊，它会睡着并变成“肉”。此时剩下 $N-1$ 只老虎面对一块肉。

由归纳可得：

• 若 $N$ 为偶数，则 $N-1$ 为奇数，剩余老虎中会有人愿意去吃这块“肉”，于是先吃羚羊的老虎会被吃掉，因此无人敢先吃；羚羊安全。
• 若 $N$ 为奇数，则 $N-1$ 为偶数，剩余老虎都不敢吃这块“肉”，于是先吃羚羊的老虎是安全的，会有人吃羚羊。

因此羚羊能安静吃草，意味着题目情景对应 $\boxed{N\text{ 为偶数}}$（此时没有老虎会先动手）。

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英文解析

This is exactly the same type of problem as the classic "Lion and Meat" inductive puzz \leq.

If a tiger eats the antelope first, it will fall asleep and turn into "meat." At this point, $N-1$ tigers face a piece of meat.

By induction:

• If $N$ is even, then $N-1$ is odd. Among the remaining tigers, someone will be willing to eat this "meat," so the tiger that ate the antelope first will be eaten; therefore, no one dares to eat first, and the antelope is safe.
• If $N$ is odd, then $N-1$ is even. The remaining tigers will all refuse to eat this "meat," so the tiger that ate the antelope first is safe, and someone will eat the antelope.

Therefore, the antelope being able to eat grass peacefully implies that the problem scenario corresponds to $\boxed{N\text{ being even}}$ (at which point no tiger will act first).