For $\displaystyle 1 \le k \le 24$ you can definitely show that the master must have played exactly $k$ games on some set of consecutive days, using pigeonhole principle.
Suppose the total number of games the master has played till the end of day $\displaystyle j$ is $\displaystyle g_j$.
Now consider the $\displaystyle 154$ numbers: $\displaystyle g_1, g_2, \dots, g_{77}, g_1 + k, g_2 + k, \dots, g_{77} + k$
These are a set of $\displaystyle 154$ numbers between $\displaystyle 1$ and $\displaystyle 132+k$.
For $\displaystyle k \lt 22$, two of these must be equal. Since $\displaystyle g_i \neq g_j$ (at least one game a day) we must have that $\displaystyle g_j + k = g_i$ for some $\displaystyle i,j$.
For $\displaystyle k=22$ we must have that the numbers are $\displaystyle 1,2, \dots, 154$, in which case, the first (and last) $\displaystyle 22$ days, the master must have played $\displaystyle 1$ game everyday.
For $\displaystyle k=23$, we can assume $\displaystyle g_i \neq 23$, and since $g_i \ge 1$, we have $g_i + k \neq 23$.
Thus by an argument similar to above, we must have have $\displaystyle 154$ numbers taking all values in $\displaystyle 1, 2, \dots, 155$, except $\displaystyle 23$ and the master must have played $\displaystyle 1$ game each of the last $\displaystyle 23$ days.
For $\displaystyle k=24$, you can show that the master must have played $\displaystyle 1$ game the first $\displaystyle 23$ days (after eliminating one of the numbers in $\displaystyle 133, 134 \dots$), then a big number of games the next, violating the $\displaystyle 12$ games per week restriction (this is where we actually used that restriction for a specific week).
We might be able to use this kind of argument to show for $\displaystyle k$ close to $\displaystyle 24$, but for larger $\displaystyle k$, I am guessing that you can find a set of games which will miss that (perhaps a computer search will help there).
Let $r_i$ be the number of sets played up to and including day $i$. So
$0 = r_0 < r_1 < r_2 < ... < r_{21} \leq 36$
So there are 37 pigeonholes - possible values - for the 22 pigeons - the $r_i$.
However, we are looking for pairs $i, j$ with $r_j = r_i + 21$ - since such a pair would tell us that 21 sets were played on days $i+1$ through to $j$. So if we assume there are no such pairs, we can "combine" the pigeonholes $i$ and $i + 21$. So we have the following pigeonholes:
$\{0 \text{ or } 21\}, \{1 \text{ or } 22\}, ..., \{15 \text{ or } 36\}, \{16\}, ... \{20\}$.
And we have 22 pigeons, 21 pigeonholes, giving a contradiction.
Best Answer
It's a bit tricky to see Pigeon Hole Principle. Here we note that the total game play is not to exceed $11\times 12=132$ games. Now, if we denote $s_i$ to be the total games played after the $i$-th day. Then look at the sequence of $\color{red}{154}$ numbers: $$s_1,s_2,\dots, s_{77}, s_{1}+21,\dots, s_{76}+21,s_{77}+21$$ Now these are the "pigeons" whereas their values are "holes". Clearly, the maximal value (i.e. the number of holes) is $132+21=\color{red}{153}$.