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:
$$d_{ij} = \begin{cases} i+j \pmod{n-1} & \text{if } i, j \neq n \\ 2i \pmod{n-1} & \text{if } i \neq n \text{ and } j = n \\ 2j \pmod{n-1} & \text{if } i = n \text{ and } j \neq n \end{cases}$$
Let us show that $d_{ij} = d_{ik}$ implies $j = k$. If $i = n$, we have $2j \equiv 2k \pmod{n-1} \Leftrightarrow j \equiv k \pmod{n-1}$. Since $j, k \le n-1$, we must have $j = k$. If $i \ne n$ and $j, k \ne n$, $i+j \equiv i+k \pmod{n-1} \Leftrightarrow j \equiv k \pmod{n-1}$, which implies $j = k$. If $i \ne n$ and $j = n$ and $k \ne n$, we have $2i \equiv i+k \pmod{n-1} \Leftrightarrow i \equiv k \pmod{n-1}$, which implies $i = k$, absurd. Hence $d_{ij} = d_{ik} \Leftrightarrow j = k$.
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.