In a round-robin tournament every two teams play each other exactly once, and every match has a…

(a) Each pair of teams plays once: $\binom{9}{2} = 36$ matches. In the figure the numbers of wins are $3, 2, 2, 2, 1$. Using the formula of (b) for $5$ teams, the number of cycles is $\binom{5}{3} - (3 + 1 + 1 + 1 + 0) = 4$. Listing them: $B, C, D$; $A, B, E$; $A, C, E$; $A, D, E$.

(b) Proof. Take any three teams. The three matches among them give three wins in all. If each team has one of these wins, the three teams form a cycle. Otherwise some team won both of its matches inside the triple, and only one team can do that, because two such teams would each have beaten the other. So every triple that is not a cycle has exactly one top team, which beat the other two. Conversely, a team together with two of the teams it beat is a triple that is not a cycle, with that team on top. Team $i$ beat $w_i$ teams, so it is the top team of exactly $\binom{w_i}{2}$ triples. Hence the number of triples that are not cycles is $\sum \binom{w_i}{2}$, and the number of cycles is $\binom{9}{3} - \sum \binom{w_i}{2}$.

(c) $\sum \binom{w_i}{2} = 28 + 15 + 10 + 10 + 6 + 3 + 1 + 1 + 0 = 74$, so there are $84 - 74 = 10$ cycles.

(d) The wins add up to the number of matches: $\sum w_i = 36$. By the Cauchy-Schwarz inequality (or the mean of squares being at least the square of the mean), $\sum w_i^2 \ge \dfrac{36^2}{9} = 144$. So $\sum \binom{w_i}{2} = \dfrac{\sum w_i^2 - \sum w_i}{2} \ge \dfrac{144 - 36}{2} = 54$, and the number of cycles is at most $84 - 54 = 30$. To reach it, number the teams $0$ to $8$ and let team $i$ beat teams $i + 1$, $i + 2$, $i + 3$ and $i + 4$, with the numbers taken modulo $9$. For two teams whose numbers differ by $d$ modulo $9$, exactly one of $d$ and $9 - d$ is in $\{1, 2, 3, 4\}$, so every match has exactly one winner. Every team wins $4$ matches, so $\sum \binom{w_i}{2} = 9 \times 6 = 54$ and there are $30$ cycles.

(e) Now there are $\binom{10}{3} = 120$ triples and $\sum w_i = 45$. The same Cauchy-Schwarz step gives only $\sum w_i^2 \ge 202.5$ and a bound of $41.25$ cycles, which cannot be reached, because the $w_i$ are whole numbers. Instead, suppose two of the numbers satisfy $u \ge v + 2$. Replacing them by $u - 1$ and $v + 1$ keeps the sum and changes $\binom{u}{2} + \binom{v}{2}$ by $-(u - 1) + v \le -1$. So at the least value of $\sum \binom{w_i}{2}$ over whole numbers with sum $45$, no two of the numbers differ by $2$ or more. Ten whole numbers with sum $45$ that differ by at most $1$ must be five numbers $4$ and five numbers $5$, giving $5 \times 6 + 5 \times 10 = 80$. Hence there are at most $120 - 80 = 40$ cycles. To reach it, take the tournament of (d) and add a tenth team that beats teams $0$ to $4$ and loses to teams $5$ to $8$. Teams $0$ to $4$ keep $4$ wins, teams $5$ to $8$ now have $5$, and the new team has $5$. So the greatest number is $40$.