1. On an island, there live knights who always tell the truth, and liars who always lie. On the main holiday, 100 islanders sat around a large round table. Half of those present said: "both my neighbors are liars," and the rest said: "one of my neighbors is a liar." What is the maximum number of knights that can sit at this table?
# Answer: 67

Solution. Note that three knights cannot sit in a row, since then the middle knight would not be able to utter any of the required phrases. Let $k$ be the number of pairs of neighboring knights. Then each of the knights said the phrase "among my neighbors there is exactly one liar," so $2 k \leqslant 50$. Moreover, to the left of this pair there is definitely a liar, and all these liars are different. Let's call such two knights and a liar a triplet. In triplets, $3 k$ islanders are involved. Consider the remaining $100-3 k$ islanders. No two knights among them sit next to each other, nor next to knights from the triplets (otherwise, three knights in a row would form, which is impossible). Then to the left of each knight, there must be a liar, all these liars are different and distinct from the liars included in the triplets. Thus, the number of knights is no more than $2 k+\frac{1}{2}(100-3 k)=50+\frac{k}{2} \leqslant 50+\frac{25}{2}=67 \frac{1}{2}$. Therefore, the number of knights is no more than 67.

Here is an example showing that 67 knights can be present:

$$
\text { LKL LKL LKL ... LKL LK LK LK ... LK L }
$$

(the block "LKL" is repeated 25 times, the block "LK" - 12 times).