XXI OBM

First case: $n$ is even. Let $T_1,T_2,\dots ,T_n$ be the teams. Each team shall play $n-1$ times on at least $n-1$ Sundays. Let us show that $n-1$ Sundays are sufficient to set up the championship.

Teams $T_i$ and $T_j$ play on the $d_{ij}$-th Sunday. Define $d_{ij}$ by the following rules: $$d_{ij}=\{\begin{array}{ll}i+j(\mathrm{mod}n-1)& \text{if }i,j\ne n\\ 2i(\mathrm{mod}n-1)& \text{if }i\ne n\text{ and }j=n\\ 2j(\mathrm{mod}n-1)& \text{if }i=n\text{ and }j\ne n\end{array}$$ Let us show that $d_{ij}=d_{ik}$ implies $j=k$. If $i=n$, we have $2j\equiv 2k(\mathrm{mod}n-1)\iff j\equiv k(\mathrm{mod}n-1)$. Since $j,k\le n-1$, we must have $j=k$. If $i\ne n$ and $j,k\ne n$, $i+j\equiv i+k(\mathrm{mod}n-1)\iff j\equiv k(\mathrm{mod}n-1)$, which implies $j=k$. If $i\ne n$ and $j=n$ and $k\ne n$, we have $2i\equiv i+k(\mathrm{mod}n-1)\iff i\equiv k(\mathrm{mod}n-1)$, which implies $i=k$, absurd. Hence $d_{ij}=d_{ik}\iff j=k$. Thus $n-1$ Sundays are sufficient to set up the championship.

Second case: $n$ is odd. Each team must play $n-1$ times on at least $n-1$ Sundays. However on each Sunday at least one team does not play. Hence $m\ge n$. Let us prove that $n$ Sundays are sufficient. Let $T_0$ be a "dummy" team. If $T_i$ plays against $T_0$ on a given Sunday, it means that $T_i$ doesn't play on that Sunday. Thus we have $n+1$ teams. As $n+1$ is even, as shown above, we know that $(n+1)-1=n$ Sundays are sufficient.