There were $2n,$ $n\ge2,$ teams in a tournament. Each team played against every other team once without draws. A team gets 0 points for a loss and gets as many points for a win as its current number of losses. For which $n$ all the teams could end up with the same non-zero number of points?
Problem
Source: Kyiv mathematical festival 2019
Tags: Kyiv mathematical festival, combinatorics
13.05.2019 09:31
Clarification: the result of each team is a string of letters "W" and "L". For example, string "WWLWLLLWLW" gives 0+0+1+4+5 points.
13.05.2019 16:50
I am interested for the case when $n$ is odd. Answer: any integer $n\geq 2$. We show by induction that for any $n$, we can set the patterns of $n$ teams be $WLWL\cdots LW$ and other $n$ teams be $LWLW\cdots WL$. If so, all team's points are positive and same except $n=1$. For $n=1$ it is obvious. If we have shown for $n$ and let $s_1,s_2,\ldots, s_n$ be teams with pattern $WLWL\cdots LW$ and $t_1,t_2,\ldots, t_n$ be teams with pattern $LWLW\cdots WL$. Now, we are given new teams, say $s_{n+1}$ and $t_{n+1}$, and set the matches between new teams and odld teams after original match. First we hold matches for which $s_{n+1}$ wins against $s_1$, $s_{n+1}$ loses against $t_1$, $t_{n+1}$ loses against $s_1$ and $t_{n+1}$ wins against $t_1$. Then, the added pattern for $s_{n+1}$ is $WL$, $t_{n+1}$ is $LW$, $s_1$ is $LW$ and $t_1$ is $WL$. Similarly, we set other matches for $s_2$, $t_2$, $s_3$, $t_3$ and so on. Finally, we set a match for which $s_{n+1}$ wins against $t_{n+1}$. Then, we get the configuration for $n+1$, completing induction.
03.12.2024 16:27
It is possible for any integer n for $X_1, X_2, \cdot \cdot \cdot X_2n $, and let $L$ for lose and $W$ for win, $LWLWLWLW \cdot\cdot\cdot L$ and $WLWLWLWL \cdot \cdot \cdot W$ will have the same score. So, just let $X_1,X_3, \cdot\cdot\cdot X_2n-1$ $LWLWLWLW \cdot\cdot\cdot L$ and the others the opposite. Then, all the players would win $ \frac{n(n+1)}{2}$ .