# Answer: 7.

Solution. After passing the coins, each person sitting at the table can have 0, 1, or 2 coins. The total number of coins will be 10. Note that if a person lies, they will state a number of coins that differs from the actual number by 1 or 2. Since the total number of coins based on the answers differs from the actual total by $10-5=5$, at least 3 people must have lied. Therefore, there are no more than 7 knights at the table. Suppose the knights sitting at the table pass the coins as follows (arrows indicate the direction of the coin transfer; the number of coins after the transfer is in parentheses): $\leftarrow \mathrm{P}(0)-\mathrm{P}(0) \rightarrow \mathrm{P}(1) \rightarrow \mathrm{P}(1) \rightarrow \mathrm{P}(1) \rightarrow \mathrm{P}(1) \rightarrow \mathrm{P}(1) \rightarrow \mathrm{L}(1) \rightarrow$ $\mathrm{L}(2) \leftrightarrow \mathrm{L}(2) \leftarrow$. In this case, all knights tell the truth, while the liars lie, saying they have 0 coins.

Comment. It is proven that there are no more than 7 knights at the table - 5 points. It is proven that there can be 7 knights at the table - 2 points.
7.5. A round table was sat by 10 people - liars and knights. Liars always lie, while knights always tell the truth. Each of them was given a coin. Then each of those sitting passed their coin to one of their two neighbors. After that, 5 people said: “I have one coin,” while the other 5 said: “I have no coins.” What is the maximum number of knights that could have been sitting at the table?