10.4. Ten chess players over nine days played a full round-robin tournament, during which each of them played exactly one game with each other. Each day, exactly five games were played, with each chess player involved in exactly one of them. For what maximum $n \leq 9$ can it be claimed that, regardless of the schedule, at the end of some game day with number $k \leq 8$, there will definitely be $n$ chess players who have already played all their scheduled games with each other?
Answer. $n=5$.

Solution. By the end of the eighth day, each chess player has played 8 games, meaning they have not played one. The unplayed games divide the chess players into 5 non-intersecting pairs. By the Pigeonhole Principle, among any six chess players, there will always be two belonging to the same pair, that is, who have not played against each other even by the end of the penultimate day of the tournament. Therefore, $n \leq 5$.

We will show that by the end of the seventh day of the tournament, regardless of the schedule, there will always be five chess players who have already played all their scheduled games with each other, from which it follows that $n=5$. We will construct a chain of chess players: the first is any player, the second is the one who played the first on the eighth day, the third is the one who played the second on the ninth day, the fourth is the one who played the third on the eighth day, and so on. Since the number of chess players is finite, at some point in the chain, a repetition will occur for the first time. This cannot happen at any player except the first, otherwise, the repeated player would have played at least three games in two days. Therefore, the chain will form a cycle of even length, as the days of the games considered alternate when moving along it. If the constructed cycle includes all 10 chess players, we stop; if not, we choose any player not included in it and repeat the procedure for constructing the cycle. The new cycle will not intersect with the already constructed one. After several steps, we will construct several cycles of even length, not less than 4, including all 10 chess players (there are few options - either 10 or $6+4$). We select half of the participants from each cycle, every other one, totaling 5 chess players. No two of them are adjacent in the cycles, so they did not play against each other on the eighth and ninth days.

Therefore, all games between them were played in the first 7 days. Thus, $n \geq 5$, and hence $n=5$. In this case, $k=7$.

Grading criteria. ($\cdot$) Proof that $n \leq 5: 3$ points. ($\cdot$) Proof that $n \geq 5: 4$ points.

10.5.. Find the maximum value of the sum $x_{1}\left(1-x_{2}\right)+x_{2}\left(1-x_{3}\right)+\ldots+x_{6}\left(1-x_{7}\right)+x_{7}\left(1-x_{1}\right)$ for arbitrary numbers $x_{1}, x_{2}, \ldots, x_{7}$ from the interval $[0,1]$.

Answer. 3.

Solution. We isolate in the left-hand side of the inequality all terms containing the variable $x_{1}$, obtaining $x_{1}\left(1-x_{2}-x_{7}\right)$. For fixed other variables, this expression, and hence the entire left-hand side, reaches its maximum when $x_{1}=1$ if $x_{2}+x_{7}<1$ and when $x_{1}=0$ if $x_{2}+x_{7} \geq 1$. Therefore, if we replace each variable, starting from $x_{1}$ to $x_{7}$, with 1 or 0 according to the condition whether the sum of the neighboring variables is less than 1 or not less than 1, we will not decrease the value of the sum on the right-hand side of the inequality. Thus, the maximum of the right-hand side is achieved for a set of $x_{1}, x_{2}, \ldots, x_{7}$ where each variable is either 1 or 0. Moreover, in such a set, two adjacent variables cannot both be 1, as in this case, the second one could not have appeared according to the replacement rules. Therefore, the set contains no more than 3 ones, and the rest are zeros, meaning that no more than three terms in the right-hand side of the inequality are non-zero, each obviously not exceeding 1, and the entire sum is no more than 3. An example where the value 3 is achieved: $x_{1}=x_{3}=x_{5}=1$, the rest are 0.

Grading criteria. Proven that the maximum of the sum does not exceed 3: 5 points. Example where the value 3 is achieved: 2 points.