218. In the World and European championships, 20 teams participate. Among them, there are $k$ European teams, the results of whose matches in the World Championship count towards the European Championship. The championship is played in a single round-robin format.

What is the largest $k$ for which it can happen that the team that scores the strictly highest number of points in the European Championship scores the strictly lowest number of points in the World Championship, if this is:

a) a hockey championship (ties are allowed)?

b) a volleyball championship (no ties allowed)?
218. Let's consider the case of $n$ participating teams in the World Championship immediately.

c) Answer: $k=n-2$, if $n \geqslant 3$ (in particular, for $n=20$ we get $k=18$).

The total number of points played in the World Championship (2 points for a win, 1 point for a draw, 0 points for a loss) is $n(n-1)$, and in the European Championship it is $-k(k-1)$.

Let $x$ be the number of points scored by the European champion in the World Championship games, and $y$ be the number of points scored by him in games with European teams, where $x \geqslant y$. Suppose each of the other teams scored more than $x$ points in the World Championship games, i.e., at least $x+1$ points. Then $\left(k+(n-1)(x+1) \leqslant n(n-1)\right)$, or $x \leqslant n-2+\frac{1}{n}$, and since $x$ is an integer, $x \leq n-2$.

Each of the other European teams scored at least $y-1$ points in the European Championship and therefore $y+(y-1)(k-1) \geqslant k(k-1)$, or $y \geqslant k-\frac{1}{k}$, and since $y$ is an integer, $y \geqslant k$. Thus, $k \leqslant y \leqslant x \leqslant n-2$, i.e., $k \leqslant n-2$.
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Fig. 90. a) Ice hockey championship; $n$ teams, $k=n-2$. Teams $3, \ldots, n$ are European, team 3 is the European champion. b) Volleyball championship; $n=8, k=3$. Teams $6,7,8$ are European, team 6 is the European champion. Teams A and B are added. c) Volleyball championship; $n=7, k=3$. Teams 5, 6, 7 are European, team 5 is the European champion. Teams A and B are added. d) Volleyball championship; $n=6, k=2$. e) Volleyball championship; $n=5, k=2$.

In Fig. 90a, an example of a tournament table is given, showing that the European champion (the third team) can occupy the last place in the World Championship.

6) Answer: $k=n-5$ for even $n \geqslant 8$ (in particular, $k=15$ for $n=20$), $k=n-4$ for odd $n \geqslant 7$; $k=2$ for $n=6$ and $n=5$.

In the volleyball championship, the winning team gets 1 point, the losing team gets 0 points, so exactly $n(n-1) / 2$ points are played. Let's consider in detail the case of even $n=2l$.

Reasoning as above, we get ($x, y$ are the same as in case a))

\[
\begin{gathered}
x+(2l-1)(x+1) \leqslant 2l(2l-1) / 2 = l(2l-1) \\
y+(k-1)(y-1) \geqslant k(k-1) / 2
\end{gathered}
\]

From equation (1), it follows that $x \leqslant l-\frac{3}{2}+\frac{1}{2l}$. Therefore, $x \leqslant l-2$. From the inequalities $l-2 \geqslant x \geqslant y$ and inequality (2), it follows that

\[
k^2 + (5-2l)k - 2 \leqslant 0
\]

For $l \geqslant 4$, the largest integer $k$ satisfying inequality (3) is $2l-5$ (for $k=2l-5$, inequality (3) holds, and for $k=2l-4$, its left side is positive).

Now let's prove that the equality $k=2l-5=n-5$ for $l \geqslant 4$ is possible. In Fig. 90c, a tournament table is given for $n=8, k=3$ (the last 3 teams are European). This table will serve as the base of induction.

Further constructions are carried out by induction, adding 2 European teams at each step.

First, note that for $k=2l-5$, it is necessarily $x=y=l-2$. Indeed,

$y \geqslant \frac{(k-1)(k+2)}{2k} = \frac{k}{2} + \frac{1}{2} - \frac{1}{2k}$, or $y \geqslant l-2 - \frac{1}{2(2l-5)}$,

and since $y$ is an integer, $y = l-2 = x$. This means that the European champion lost to all non-European teams.

Now suppose that a tournament table with $k=n-5=2l-5$ has already been realized. Add two European teams $A$ and $B$ according to the following rule: team $A$ wins against the old teams with odd numbers and against team $B$ and loses to the rest; team $B$ wins against the teams with even numbers and loses to the rest. Each of the previous teams receives one point as a result, so their relative positions do not change.

The European champion now has $l-1$ points, team $A$ has $l-1$ points, and team $B$ has $l$ points, so the old European champion remains in the last place. In games with European teams, team $A$ will score $l-2$ points ($l-3$ in games with "old" teams and one with team $B$); team $B$ will also score $l-2$ points in games with European teams, so the European champion remains the same.

The cases of odd $n \equiv 2l-1$ for $l \geqslant 4$, as well as $n \leqslant 0$, are considered similarly (Fig. 90d, e).