# Solution.

According to the condition, an even number of teams $2n$ started, of which $n$ were disqualified. The eliminated teams played $\frac{n(n-1)}{2}$ matches among themselves. The same number of matches were played by the teams that remained in the tournament.

Let each of the eliminated teams have played with $k$ teams from among the remaining ones $(k>1$. Clearly, $k=1$ is not a solution, so $n+k-1>n$. Factoring $77=7 \cdot 11$, we get

$$
n=7, \quad n+k-1=11
$$

from which $k=5$.

## Answer: 14 teams.
# Problem 2.

In a football tournament, each team is supposed to play one match against each of the others. But during the tournament, half of all the teams were disqualified and withdrew from further participation. As a result, 77 matches were played, and the teams that withdrew had managed to play all their matches among themselves, with each of the withdrawn teams playing the same number of matches. How many teams were there at the beginning of the tournament?