I'm posting this answer to my own question because I think I found a solution, but I would appreciate feedback on whether the solution below is complete.
The only detail that is missing to complete part d is that I need to show that the statement is true if 24 in the original statement is replaced by 30; that is, I will try to show that there is a period of consecutive hours during which the arm wrestler had exactly 30 matches (since there seems to be no counter-example for this case).
I think I found a way to prove it, based on the suggestion given by Lopsy in the comments to the question. This attempt is very similar to the accepted answer here: Prove that 2 students live exactly five houses apart if.
We build the following sets: the 5-element sets $\{1, 31, 61, 91, 121\}$, $\{2, 32, 62, 92, 122\}$, ..., $\{5, 35, 65, 95, 125\}$, and the 4-element sets $\{6, 36, 66, 96\}$, $\{7, 37, 67, 97\}$, ..., $\{30, 60, 90, 120\}$. The purpose of these sets is to represent every possibility of number of matches played so far. They encompass all numbers from 1 to 125 and, inside each set, the difference between adjacent numbers is 30; however, there are no numbers from two different sets whose difference is 30. There are 5 five-element sets, and 25 four-element sets.
Now, if I want to choose 75 numbers from all sets so that there are no two numbers whose difference is 30, I have to choose at most 3 elements from the sets with 5 elements (for example, I can choose 1, 61 and 121 from the set $\{1, 31, 61, 91, 121\}$; if I choose one more, then clearly two of them will have difference equal to 30), and at most 2 elements from the sets with 4 elements (for example, I can choose 30 and 120 from $\{30, 60, 90, 120\}$, but, if I choose one more, then clearly two of them will have difference 30). There are 30 sets in total; since there are 5 sets with 5 elements and 25 sets with 4 elements, the maximum number of elements that can be chosen so that no two of them have 30 as difference is $3\times 5 + 2\times 25 = 15 + 50 = 65$ numbers. But I need to choose 75 numbers. So, there must be some numbers whose difference is 30.
Does this make sense, or is there some mistake?
Best Answer
Suppose that the sums of sequences of three adjacent numbers in the circle are $s_1,s_2,\dots,s_{20}$. When you form the grand sum $s_1+s_2+\cdots+s_{20}$, in effect you’re adding up the numbers from $1$ through $20$ three times (why?), so you know the total. If all of the $s_k$ were less than $32$, what would the maximum possible total be?