An American college football championship is held in the following system: First, the teams play round-robin matches in groups of five or six. Then comes the knockout stage, where only those teams advance that have suffered no more than one defeat in the group matches. How many teams can advance from each group?
We solve the task more generally: suppose that the groups consist of $k$ teams, and those who suffer at most $m$ defeats advance. The question is the same as the original: how many can advance from a group? When $m \geq k-1$, of course everyone advances, so in the following we assume that $m \leq k-2$ and accordingly $k \geq 3$.

Let the number of qualifiers be $t$, and first examine the matches they played against each other. It can be seen that this represents $\frac{t(t-1)}{2}$ matches: everyone played with $t-1$ others, which is $t(t-1)$, but we counted everything exactly twice. Every match had a loser, so the $t$ qualifying teams suffered at least $\frac{t(t-1)}{2}$ defeats in total. None of them lost more than $m$ times, so

$$
\frac{t(t-1)}{2} \leq t m
$$

holds. This is clearly true for $t=0$, otherwise it can be simplified to

$$
t \leq 2 m+1
$$

Since this also holds for $t=0$, we have thus found that the number of qualifiers can be at most $2 m+1$.

We apply a similar consideration to the matches between the eliminated teams. Each of them could have lost to at most $t$ qualifiers, and since they lost at least $m+1$ times in total (since they were eliminated), they must have lost at least $m-t+1$ times to other eliminated teams. Thus, the eliminated teams lost at least $(k-t)(m-t+1)$ times to each other. This cannot be more than the total number of matches they played against each other, i.e.,

$$
\frac{(k-t)(k-t-1)}{2} \geq(k-t)(m-t+1)
$$

In the case of $k \neq t$ (since then $k > t$, as $k \geq t$ is obvious),

$$
k-t-1 \geq 2 m-2 t+2, t \geq 2 m-k+3
$$

This also includes the case $k=t$ (based on $k-2 \geq m$).

Thus, we have found that for the possible $t$ numbers,

$$
2 m-k+3 \leq t \leq 2 m+1
$$

Now we show that these are indeed valid: for every such $t$, we give the outcome of the matches such that $t$ teams advance. For this, we first select $t$ teams - they will be the qualifiers, the others will be eliminated. We also choose the outcome of the matches among them so that the qualifying team always wins.

Now let's examine the matches between the qualifiers. Consider first the case $t=2 m+1$. We arrange the teams in a circle; everyone beats the $m$ teams to their right and loses to the others. This ensures that these $t$ teams will indeed advance. If $t > k$ or $2 m-k+3 < 0$ holds, and such $t$ values are clearly not possible. Therefore, at most the

$$
\max (0,2 m-k+3) \leq t \leq \min (k, 2 m+1)
$$

statement can be true. However, then problems arise in the construction: when we complete the $t$ qualifiers to $2 m+1$, perhaps there are not enough teams, and the same may apply to the completion of the eliminated teams to $2 k-m-3$. However, the following consideration can eliminate these problems.

If more teams need to be added than are available, then we also use "fictitious" teams. We distribute the match outcomes in the same way, then first we remove the fictitious teams (along with their matches), and then the others (correcting their matches). This operation does not increase the number of losses of the $t$ teams selected for advancement, and for the eliminated ones, it does not increase their wins. That is, the $t$ teams selected for advancement will still have at most $m$ losses after removal, and the other $k-t$ will have at most $(k-1)-(m-1)$ wins, so the $t$ teams will advance, and the others will not.

Thus, we have given a valid outcome of the tournament for every $t$, thereby solving the problem. For the cases $k=5, m=1$ and $k=6, m=1$, the number of qualifiers satisfies

$$
\max (0,2-5+3)=0 \leq t \leq 3, \quad \text { and } \quad \max (0,2-6+3) \leq t \leq 3
$$

respectively, so 0, 1, 2, or 3 teams can advance.

Based on the work of Gyula Pap (Debrecen, Fazekas M. Gymnasium, 2nd year)

Note. Many solvers only determined the maximum number of qualifiers, so they received partial points. Those who only sent a diagram did not receive points. Of course, those who solved the original problem for the specific case (with a similar line of thought as the given solution) received 5 points.