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.
This is not a solution to the current problem.
Using the standard setup, I will show that
- With at most 13 tasks per calander week, over 15 weeks, then there is a series of days with exactly 33 tasks.
- With at most 12 tasks per calander week, over 13 weeks, then there is a series of days with exactly 33 tasks.
In either case, this standard setup doesn't allow us to reduce the number of weeks.
I suspect the following statement is true (and possibly something stronger), as I cannot find a counter example:
3. With at most 13 tasks per calander week, over 13 weeks, then there is a series of days with exactly 33 tasks.
Note: Obviously if we're allowed 14 tasks per week, then we could do 2 a day and never get a series of days with exactly 33 tasks.
Proof of 1:
Let $t_i$ be the number of tasks done on day $i$.
Let $T_i = \sum_{j=1}^i t_j $ be the cumulative number of tasks done by day $i$.
We have $1 \leq T_1 < T_2 < \ldots < T_{105} \leq 195$.
Let our pigeons be $T_i$. There are 105 of them.
Let our pigeonholes be the sets of the form $\{ 66k + i, 66k+i+33 \}$. Since $66 \times 3 = 198 > 195$, there are 99 of them.
So, by PP, there are 2 piegons in 1 hole, which gives us $T_j = T_i + 33$.
Proof of 2.
Set up in a similar manner.
We have $1 \leq T_1 < T_2 < \ldots < T_{91} \leq 156$.
Let our pigeons be $T_i$. There are 91 of them.
Let our pigeonholes be the sets of the form $\{ 66k + i, 66k+i+33 \}$. Since $156 = 2\times 66 + 24$, there are $66 + 24 = 90$ of them.
So, by PP, there are 2 piegons in 1 hole, which gives us $T_j = T_i + 33$.
Thoughts on 3.
There are 91 pigeons and 99 holes.
We have 8 degrees of freedom in choosing the values of $T_i$.
Best Answer
Let $E$ be the set $\{x_1,x_1+x_2,x_1+x_2+x_3,\ldots,x_1+x_2+\ldots+x_{13}\}$. $E$ is made by $13$ elements in $[1,20]$.
$E+4$ is made by $13$ elements in $[5,24]$, so if we assume that $E$ and $E+4$ are disjoint we get $$ 26=|E|+|E+4|=|E\cup(E+4)|\leq |[1,24]| = 24, $$ a contradiction. It follows that two elements of $E$ differ by $4$.