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).
Call the sequence $a_1,\ldots,a_m$ and the $n$ distinct terms $t_1,\ldots,t_n$. For each $k$ from $1$ to $m$ define the set
$$S(k)=\{j\mid \hbox{$t_j$ occurs an odd number of times among $a_1,\ldots,a_k$}\}\ .$$
Now consider two cases.
For some $k$ we have $S(k)=\varnothing$.
$S(k)$ is never empty. Then there are fewer than $m$ distinct $S(k)$ so at some point we have $S(k_1)=S(k_2)$ and then...
Since you asked for hints rather than a solution I'll leave it there....
Best Answer
For each day, $\binom52=10$ of the $\binom{20}2=190$ possible pairs of jewels will be worn. Over $267$ days, this makes $2670$ pairs worn, but $\frac{2670}{190}=14.05\ldots>14$. This means that at least one pair must have been worn at least $15$ times.
The holes are the possible pairings of jewels and the pigeons are the pairs actually worn together on some day.