What is the minimum wins needed to be in top 50 percent of teams in a round robin tournament? Consider a championship among $2n$ teams. The first round of the championship consists of a round robin tournament among all $2n$ teams. The top $n$ teams progress to the next round. Ties are resolved (by coin toss or some other parameter) only when multiple teams are fighting for the $n^{th}$ slot. 
How many wins would be "sufficient" for a team to guarantee that it reaches the next round of the championship?
Alternatively, what are the number of wins that a team needs, so that it doesn't need to worry about the results of the games between the rest of $2n-1$ teams?
My intuition and observations:-
A semi-regular tournament on $2n$ vertices would have $n$ vertices with $n-1$ outdegree(wins) and $n$ vertices with $n$ outdegree(wins).
Thus $n-1$ wins is not sufficient.
I guess that any team with $\geq (n+1)$ wins is guaranteed a slot in the next round of the championship?
 A: When $n+1$ is not sufficient, we'll have at least $n+1$ teams with score of at least $n+1$. That means there will have been a total of at least $(n+1)^2$ games, so we need
$$(n+1)^2 \leqslant \binom{2n}2 \implies n\geqslant 4.$$
Already for $n=4$ we can find one such configuration.
Choose $5$ of the teams and have each of them beat all of the other $3$.
Additionally, arrange these $5$ chosen teams in an oriented circle and have each of them beat the next $2$.
In this way, the $5$ chosen teams will each have a score of $5$, while the other $3$ teams will each have a score of at most $2$.

More generally, this can be adapted to show that no given $n+k$ works when $k$ is independent of $n$.
For a given $k$, let $n$ be sufficiently large.
We've already seen that we need $(n+k)^2 \leqslant \binom{2n}2$, but we'll see that the circle arrangement demands another condition.
Choose $n+k$ of the teams and have each of them beat all of the other $n-k$.
Additionally, arrange these $n+k$ chosen teams in an oriented circle and have each of them beat the next $2k$.
This requires that
$$2k \leqslant \frac{(n+k)-1}2 \iff 3k+1 \leqslant n$$
This ensures that $n+k$ of the teams will each have a score of at least $n+k$, while the other $n-k$ will have a score of at most $n-k-1$.

I'll try and make im_so_meta's answer more precise.

Claim: The minimum number of wins is $\lfloor 3n/2\rfloor$.

Proof: First, suppose that achieving a score of $\lfloor 3n/2\rfloor$ did not guarantee that a team will be above the $50$th percentile.
Then there would be some win-loss configuration for which $n+1$ teams each had a score of $\lfloor 3n/2\rfloor$ or more.
All of these wins would have taken at least
$$(n+1)\left(\frac{3n}2-\frac12\right) = \frac{3n^2+2n-1}2$$
games.
The other $n-1$ teams will have played another
$$\binom{n-1}2 = \frac{n^2-3n+2}2$$
games amongst themselves, for a total of $2n^2 - (n-1)/2$.
But this is more than $\binom{2n}2 = 2n^2 - n$, which is the total number of games, so no such configuration exists.
It follows that achieving a score of $\lfloor 3n/2\rfloor$ does guarantee that a team will be above the $50$th percentile.
We now show that it's possible to achieve $\lfloor 3n/2\rfloor - 1$ without being above the $50$th percentile.
Case $(1):$ $n$ is even
In this case, $\lfloor 3n/2\rfloor - 1 = 3n/2 - 1$.
Write this as $(n-1) + n/2$.
Choose $n+1$ of the teams and have each of them beat all of the other $n-1$.
Additionally, arrange these $n+1$ chosen teams in an oriented circle and have each of them beat the next $n/2$.
Notice that here there is no freedom, and this completely determines the games amongst these teams.
This ensures that $n+1$ of the teams will each have a score of exactly $(n-1) + n/2$, while the other $n-1$ teams will each have a score of at most $n-2$.
Case $(2):$ $n$ is odd
In this case, $\lfloor 3n/2\rfloor - 1 = 3n/2 -1/2 -1$.
Write this as $(n-1) + (n-1)/2$.
Choose $n+1$ of the teams and have each of them beat all of the other $n-1$.
Additionally, arrange these $n+1$ chosen teams in an oriented circle and have each of them beat the next $(n-1)/2$.
In this case, there is freedom to choose the outcome of diametrically opposite teams arranged in the circle.
This ensures that $n+1$ of the teams will each have a score of at least $(n-1) + (n-1)/2$, while the other $n-1$ teams will each have a score of at most $n-2$.
A: @Fimpellizieri has already established $n+1$ is insufficient.
I claim that the minimum number of wins is $\frac{3n}{2}$. Assume for contradiction $n+1$ teams have this win total, then we already have $\frac{3n^2+3n}{2}$ wins logged in total by these teams, and the other $n-1$ teams must have at least $\binom{n-1}{2}=\frac{n^2-3n+2}{2}$ wins between them, for a total of $2n^2+1$. However, there are only $\binom{2n}{2} = 2n^2-n$ total wins! Oops. Contradiction.
Construction:
Have $n+1$ of the people all beat the other $n-1$ people. Now, we can have all of the $n+1$ people have $\frac{n+1}{2}$ wins amongst themselves by putting them on a circle and having them beat the next $\frac{n+1}{2}$ people in a certain direction, plus or minus a half (half of the people have plus, half have minus) (credit to @Fimpellizieri for the idea). For those who have minus a half, then we have $\frac{3n}{2}-1$ wins, and all of these people tie.
I realize this answer glosses over the cases whether $n$ is even or odd, but the answer is either $\frac{3n}{2}$ or $\frac{3n}{2}-\frac{1}{2}$.
Thanks for @Fimpellizieri for pointing out that I accidentally found the equality case, botched the arithmetic, and somehow thought it was the answer.
