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
Let $m_i$ be the number of matches fought in hour $j$, so we have $m_i \ge 1$ for $1 \le i \le 75$. Define $$s_n = \sum_{i=1}^n m_i$$ for $1 \le n \le 75$. If we consider the values $s_n$ modulo $24$, there are $24$ possible slots and $75$ numbers, so there must be some slot that contains at least $4$ numbers, by the pigeonhole principle. Let's say the $4$ numbers are $s_a, s_b, s_c$ and $s_d$, with $a<b<c<d$, so $s_a=s_b=s_c=s_d \pmod{24}$. Then $s_b-s_a = s_c-s_b=s_d-s_c = 0 \pmod{24}$, so $$\sum_{i=a+1}^b m_i = \sum_{i=b+1}^c m_i= \sum_{c+1}^d m_i = 0 \pmod{24} \tag{*}$$ Therefore each one of the three sums above must be one of the values $0, 24, 48, 72 \dots$ etc.
Zero is ruled out as a sum because we know $m_i \ge 1$ for all $i$. Can all three sums be $48$ or greater? No, because then the total of the three sums would be at least $144$, and we know the total number of matches was no more than $125$. So at least one of the sums listed in $(*)$ is equal to $24$, i.e. exactly $24$ matches were fought in one of the intervals $a+1$ to $b$, $b+1$ to $c$, or $c+1$ to $d$.