This "game" is a strengthening of the union-closed sets conjecture (background here and here).
Given a positive integer $n \ge 2$ and $m = \lfloor (n-1)/2 \rfloor$, the game starts disposing all the unordered couples in $\binom{[n]}{2}$ in a triangle as follows:
$$ \begin{array}{l} \{1,2\} & \{1,3\} & \ldots & \{1,n-1\} & \{1,n\} \\ \{2,3\} & \{2,4\} & \ldots & \{2,n\} \\ \ldots & \ldots & \ldots \\ \{n-2, n-1\} & \{n-2, n\} \\ \{n-1, n\} \end{array} $$
For example for $n=5$ the triangle is this:
$$ \begin{array}{l} \{1,2\} & \{1,3\} & \{1,4\} & \{1,5\} \\ \{2,3\} & \{2,4\} & \{2,5\} \\ \{3,4\} & \{3,5\} \\ \{4,5\} \end{array} $$
The "score" of a couple $\{a,b\} \in \binom{[n]}{2}$ is the sum of the minimum number of occurrences between $a$ and $b$ for each row, excluding the same couple $\{a,b\}$. For example, for the above triangle with $n=5$, the score of $\{4,5\}$ is $1+1+1=3$, the score of $\{3,4\}$ is $1+1=2$, the score of $\{2,3\}$ is $1$, the score of $\{1,2\}$ is $0$.
The overall score of the game is the maximum couple score over all couples.
In order to win, the player has to move couples freely in another row, but only upwards, trying to get an overall score strictly less than $m$.
Is it possible to win? Can we say something about that?
For $n=5$ it is not possible: we need to move two couples from the second row to the first, WLOG $\{2,3\}$ and $\{2,4\}$, to lower the score of $\{3,5\}$ and $\{4,5\}$, but then $\{3,4\}$ will have a score of $2=m$ all from the first row.
For $n=7,8$ I have verified that winning is impossible with an ILP.
