Real-world inspiration. My younger son was playing a micro football (soccer) tournament this afternoon with $3$ other friends. Let's label the $4$ kids $0,1,2,3$. They played $3$ matches:

1. $\{0,1\} \text{ vs } \{2,3\}$,
2. $\{0,2\} \text{ vs } \{1,3\}$,
3. $\{0,3\} \text{ vs } \{1,2\}$.

So everyone got to play in a team with everyone else exactly once. It was all splendid and satisfying, also in a mathematical way. Which led to the following.

Generalisation. Let $t, n >1$ be integers; let $t$ stand for "team size" informally, and $n$ for "number of teams". We want to split the $m := t\cdot n$ "players" into $n$ teams of $t$ players a finite number of rounds such that every player has been in the same team with every other player exactly once. It turns out that this condition -- let's call it condition (C) -- puts some restrictions on $t, n$ and $m = t\cdot n$: In every round, each player is in a team with $t-1$ other players. In the end, every player has played with the other $m-1$ players exactly once. So $t-1$ must divide $m-1$, and we get $(m-1)/(t-1)$ rounds.

So the question is: if the integers $t,n>1$ satisfy the condition that $t-1$ divides $tn -1 = m-1$, can $(m-1)/(t-1)$ rounds of $n$ teams of size $t$ be organised such that condition (C) holds? If not, what additional restrictions must hold for $t,n$?

Real-world inspiration. My younger son was playing a micro football (soccer) tournament this afternoon with $3$ other friends. Let's label the $4$ kids $0,1,2,3$. They played $3$ matches:

• $\{0,1\} \text{ vs } \{2,3\}$,
• $\{0,2\} \text{ vs } \{1,3\}$,
• $\{0,3\} \text{ vs } \{1,2\}$.

So everyone got to play in a team with everyone else exactly once. It was all splendid and satisfying, also in a mathematical way. Which led to the following.

Generalisation. Let $t, n >1$ be integers; let $t$ stand for "team size" informally, and $n$ for "number of teams". We want to split the $m := t\cdot n$ "players" into $n$ teams of $t$ players a finite number of rounds such that every player has been in the same team with every other player exactly once. It turns out that this condition -- let's call it condition (C) -- puts some restrictions on $t, n$ and $m = t\cdot n$: In every round, each player is in a team with $t-1$ other players. In the end, every player has played with the other $m-1$ players exactly once. So $t-1$ must divide $m-1$, and we get $(m-1)/(t-1)$ rounds.

So the question is: if the integers $t,n>1$ satisfy the condition that $t-1$ divides $tn -1 = m-1$, can $(m-1)/(t-1)$ rounds of $n$ teams of size $t$ be organised such that condition (C) holds? If not, what additional restrictions must hold for $t,n$?