If one plays $132$ games in $77$ days, there is a span of consecutive days with exactly $21$ games

Let $a_1, a_2, a_3, \dots, a_{77}$ be the number of games played in the $nth$ day, such that $a_i \leq 132, i\in [1, 77]$. Also consider the sequence $a_1 + 21, a_2 + 21, \dots ,a_{77} + 21$, $77 \leq a_i + 21 \leq 132 + 21 = 153$.

The 154 sequences $a_1, a_2, a_3, \dots, a_{77}$ and $a_1 + 21, a_2 + 21, \dots ,a_{77} + 21$ are all less than $153$. Thus by pigeon hole principle,

$$\Big\lceil \dfrac{154}{153}\Big\rceil = 2$$

there's at least 2 $a_j$ and $a_k + 21$, such that $a_j = a_k + 21$. We know that since $a_j$ cannot equal another $a_l$, $j \neq l$ since all the values in the sequence $a_1, a_2, a_3, \dots, a_{77}$ are distinct (at least one game a day), thus one of $a_j$ have to equal $a_k + 21$, for some $j, k \in \mathbb{Z}$. That means that there exist from $k^{th}$ day to $j^{th}$ day, there's 21 consecutive games.