10.5. The teams participating in the quiz need to answer 50 questions. The cost (in integer points) of a correct answer to each question was determined by experts after the quiz, the cost of an incorrect answer - 0 points. The final score of the team was determined by the sum of points received for correct answers. When summarizing the results, it was found that the costs of correct answers could be assigned in such a way that the teams could take places according to any wishes of the experts. What is the maximum number of teams that could have participated in the quiz
Answer: 50.

Solution: We will prove that with 50 teams, such a distribution of points can exist. The example is obvious - let the $k$-th team answer only the $k$-th question. Then, by assigning the costs of the questions as $a_{1}, a_{2}, \ldots, a_{50}$, where $\left\{a_{1}, a_{2}, \ldots, a_{50}\right\}=\{1,2, \ldots, 50\}$, the jury will place the $k$-th team at position $50+1-a_{k}$.

Suppose there are 51 teams. Imagine that we have cloned each team, meaning we have an unlimited number of teams of each of the 51 types (all teams within a type answer questions the same way). We will prove that it is possible to form two groups from them, different in composition (at least for one type, the number of teams of this type in the first group is not equal to the number of teams of this type in the second group), but having the same results (that is, on each question, the same number of teams in the first group answered as in the second).

Indeed, let's write a system of linear equations, where the $i$-th equation states that the difference in the number of teams in the first and second groups that answered the $i$-th question is zero; here $x_{j}$ is the number of teams of the $j$-th type (in the first group if $x_{j}$ is positive, in the second if negative). The coefficients - zeros and ones - are determined by whether the team of the $j$-th type answered the $i$-th question. This is a system of 50 homogeneous equations with 51 unknowns. It has a non-zero solution, and since all coefficients are rational, there exists a non-zero rational solution. Since the equations are homogeneous, the solution vector can be multiplied by a constant. Multiply it so that all $x_{j}$ values become integers. The required groups are found. In this case, teams of each type are present in no more than one group. Let the number of teams in the first group be no less than in the second. Then it is impossible to assign points to the questions in such a way that the positions of all teams in the first group are higher than the positions of the teams in the second group, because the sum of the points of the teams in the first group is always equal to the sum of the points of the teams in the second group.

In the case where there are more than 51 teams, we can consider only 51 teams, and for them, the condition of the problem does not hold.