25. a) At a sports competition, 20 athletes are participating; the competition is judged by 9 judges. Each judge ranks the competitors from 1st to 20th according to their assessment of the performances, and it turned out later that the ratings of all judges did not differ significantly: no athlete received places from two judges that differed by more than three. The final ranking is based on the "average place" of the athlete according to the ratings of the 9 judges, i.e., the sum of all assigned places divided by the number of judges (9). What is the highest possible value of the "average place" for the best of the 20 athletes?

b) The tennis federation has assigned qualification numbers to all tennis players in the country: the 1st number to the strongest of them, the 2nd to the next strongest, and so on; it is known that in a match between tennis players whose qualification numbers differ by more than two, the player with the lower number always wins. 1024 tennis players from the country are holding a competition according to the Olympic system, i.e., after each stage of the competition, all losers are eliminated from the game, and the remaining players are randomly paired for the next round. The question is, what is the highest number that the winner of such a competition can have?
25. a) It is clear that the smallest value of the "average position" is 1 - this value is achieved if all judges assign the 1st place to the same athlete. On the other hand, 5 or more athletes cannot receive (from different judges) the first place: indeed, these $n \geqslant 5$ athletes would then collectively receive 9 first places and $9n - 9 \geqslant 9 \cdot 5 - 9 = 36$ other places (since 9 judges assign them a total of $9n$ places); by the problem's condition, none of these places can be lower than 4th, which is impossible, as places from 3rd to 4th are only $3 \cdot 9 = 27$. Thus, we only need to consider the cases where 1st place (from different judges) is assigned to 2, 3, or 4 athletes.

$1^{\circ}$ If 1st place is assigned by judges to only two athletes, then one of them is named 1st by at least five judges - and since the other judges place him no lower than 4th,

the "average position" of this athlete is no less than $\frac{1}{9}(5 \cdot 1 + 4 \cdot 4) = \frac{21}{9} \left(=2 \frac{1}{3}\right)$

$2^{\circ}$ If 1st place is assigned to three athletes, then collectively these athletes receive 9 first places and another $3 \cdot 9 - 9 = 18$ other places, none of which can be lower than 4th; but since 9 judges can assign only 9 fourth places, in the "worst" case, our athletes receive 9 fourth and 9 third places. Thus, the total sum of places assigned to these three athletes is no more than $9 \cdot 1 + 9 \cdot 4 + 9 \cdot 3 = 72$, so at least one of them has a sum of places no more than $\frac{72}{3} = 24$ and an "average position" no worse than $\frac{24}{9} \left(=2 \frac{2}{3}\right)$.

$3^{\circ}$ Finally, if 1st place is assigned to four athletes, then these four athletes collectively receive 9 first places and another $4 \cdot 9 - 9 = 27$ other places, no lower than 4th; of these, 9 can be 4th places, 9 can be 3rd places, and the last 9 can be 2nd places. Thus, the total sum of places for these four athletes is $9 \cdot 1 + 9 \cdot 2 + 9 \cdot 3 + 9 \cdot 4 = 90 (< 4 \cdot 23)$; therefore, the best of these four athletes has a sum of places no more than 22 and an "average position" no worse than $\frac{22}{9} \left(=2 \frac{4}{9} < 2 \frac{2}{3}\right)$.

Thus, the "average position" of the best athlete cannot be lower than $2 \frac{2}{3}$; it can equal $2 \frac{2}{3}$ only if each of the three best athletes received 1st place from three judges, 3rd place from three others, and 4th place from the last three (so in this case, the winners of the competition will be three athletes).

Note. Of course, the same reasoning, where only the $k$-th place is replaced by the $(21-k)$-th place, shows that the "average position" of the last athlete cannot be higher than $18 \frac{1}{3}$ (but can equal $18 \frac{1}{3}$):

b) It is clear that after each round of the competition, the number of the strongest remaining participant does not decrease; it can increase by no more than 2 (if the strongest tennis player is accidentally defeated by a player whose number is two units higher than his). Since $1024 = 2^{10}$ and after each round the number of participants in the competition is halved, we will have 10 rounds of the competition, after which only $2^0 = 1$ tennis player, the winner of the competition, will remain. Since after each round the number of the best remaining player can increase by 2, it seems that the 21st player on the list could be the winner of the competition.

However, the 21st player actually has no chance of becoming the winner. Indeed, for this to happen, the two strongest players must be eliminated at each stage of the competition; in other words, the 1st and 2nd players must lose in the first round to the 3rd and 4th players, respectively; in the second round, the 3rd and 4th players must lose to the 5th and 6th players, and so on. But in this case, in the semifinals, the 17th and 18th players must be eliminated by the 19th and 20th players; therefore, the 19th and 20th players advance to the final, and the winner of the competition is one of them, not the 21st player.

Finally, we will show that the 20th player can become the winner of the competition. Indeed, with only $2^1 = 2$ players, the winner can be the player with the number $2 = 2 \cdot 1$; if there are $2^2 = 4$ players, the winner can be the player with the number $2 \cdot 2 = 4$: in the pair of the first two players, the 2nd player can win, and in the pair of the other two, the 4th player can win, and the 4th player can, in principle, win the final against the 2nd player. Similarly, if there are $2^3 = 8$ players, the 6th player can become the winner: indeed, if the four strongest players end up in one subgroup, the strongest in that subgroup can be the 4th player; in the other subgroup, the 6th player can win against the 5th player; in the final, he can meet the 4th player, whom he can defeat. In the same way, using the method of mathematical induction, it is easy to prove that in a competition of $2^n$ tennis players, the $(2n)$-th player by strength can become the winner: for this, the first $2n-2$ players must end up in one subgroup of $2^{n-1}$ players, where (by the induction hypothesis) the $(2n-2)$-th player can become the winner; in the other subgroup, the $(2n)$-th player can become the winner, and after that, he only needs to win the final against the $(2n-2)$-th player, which is a difficult but possible task.