Maximum number of players with a certain score in a tournament

Question

You have a list $L$ of $2n$ players who all start with an equal score, $k\ge 0$. In a "tournament round", we execute the following process:

1. Sort the list according to the player's scores, largest to smallest. (equality being sorted arbitrarily)
2. You partition the list into pairs, the first two players in your list are in a pair, the next two players in your list are a pair, etc.
3. In each pair, we decrease the score of one of the players by 1.

Question 1: does this process have a name?

Question 2: after $i$ iterative tournament rounds, what the maximum amount of players which can have a non-negative score? (or at least a rough bound) We can call this quantity $f(n,k,i)$.

With $$ F(n,i) := 2n(\frac12+\frac12x)^i,$$ we note that the coefficient of $x^j$ in $F(n,i)$ is roughly the number of people who lost $j$ times in $i$ tournament rounds. However once $F(n,i)$ has odd integer coefficients, we start to get a small error for larger $i$ as our coefficients will start to get fractions. I am not sure how to bound the size of this error.

Answer 1

Since this seems more of a creative problem solving question where no profound mathematical theory is needed I will go out on a limb and give an attempt of a solution (not necessarily correct!).

We have $2n$ players each with a score. Let the scores be arranged as

$$ \begin{align} k_{1}\geq k_{2} \geq k_{3} \geq \dots \geq k_{2n-1} \geq k_{2n} \geq 0, \end{align} $$

where the $j$:th player has score $k_{j}$. If the number $i<k_{2n}$ the maximum number of players with non-negative scores is $2n$ evidently. Now, let the case be that $k_{2n-1}\geq i \geq k_{2n}$, then we may decrease player $(2n-1)$:th score in each iteration such that we again obtain $2n$ players with non-negative scores. In fact, as long as the sum of the pair is greater than $i$ it is possible to decerase the scores such that they are above $0$ at the end of the process. The rule we must follow is to always decrease the best score of each pair first, then decreace the lower score. This rule will preserve the pairs in the same state, i.e. the $2k$:th player will always be paired with the $(2k-1)$:th player for $k=1,\dots,n$. Thus, the maximum number of players with non-negative scores at the end of the $i$:th tournament round is equal to the higher index of the pair with at least $i$ score in total. I hope that helped.