自指语句：100 句话里有多少句为真

设恰好有 $r$ 句为真。

那么“至多 $n-1$ 句为真”的陈述在且仅在 $n-1\ge r$（即 $n\ge r+1$）时为真。

因此为真的句子数量等于满足 $n\ge r+1$ 的句子个数，即 $100-r$。

一致性要求 $100-r=r$，所以 $r=50$。

也可验证：第 51 到第 100 句（声称“至多 50..99 句为真”）都为真，而前 50 句为假。

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Original Explanation

We know that the first statement must be false, as if it was true, it would contradict itself that none of the statements are true. Therefore, let's consider the second statement. If it was false, then this implies that at least $2$ statements are true. However, the statement after says that at most $2$ statements are true. If that were to be true, then that claims exactly $2$ statements would be true, of which one is itself. However, every statement that says at most $k > 2$ statements are true would also be true, causing a logical contradiction. Therefore, the third statement must be false.

The trick here is to notice that for some threshold $r$, all the statements saying "at least $r$ of the statements are true" must be true for everything at least that $r$, and false for everything below it. For such an $r$, that statement being true implies that both at most and at least $r$ statements are true, meaning exactly $r$ statements are true. Namely, there must be $100 - r$ statements that are false and $r$ statements that are true. To find that threshold $r$, we just equate the above, yielding $r = 50$ statements must be true. We can verify this by seeing that the $50$ statements "at most $50, 51, \dots, 99$ statements are true" must be true and other $50$ are false.