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 $d$.