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    $\begingroup$ $\binom{22}{2}=21\cdot 11$ is not a multiple of $2\binom{11}{2}=10\cdot 11$. $\endgroup$ Commented Mar 25, 2025 at 13:10
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    $\begingroup$ Suppose $n \ge 2k$. Then (by pigeonhole) in the second round there are at least $k$ students on the first team who were on a team together in round one, so we cannot even play $2$ games. $\endgroup$ Commented Mar 25, 2025 at 14:15
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    $\begingroup$ Matthew Bolan's argument also works for $n=2k-1$ $\endgroup$ Commented Mar 26, 2025 at 18:24