30th Turkish Mathematical Olympiad

Answer: $77$.

First of all, let us give an example for $77$ tours. Let us take $26\times 77=2002$ school pupils and divide them into groups $A$ and $B$ consisting of $26\times 51=1326$ and $26\times 26=676$ pupils, respectively. To each pupil from $A$ we assign a different pair from the set $$A=\{(i,j):i,j\in {\mathbb{Z}}_{+},i+j\le 52\}$$ To each pupil from $B$ we assign a different pair from the set $$B=\{(i,j):i,j\in {\mathbb{Z}}_{+},i\le 26,j\le 26\}.$$ Now for each $1\le k\le 77$ in the $k$-th day:

† if $k\le 51$ then $k$ pupils from the group $A$ for which $i=52-k$ attend a museum tour,

†† if $k\ge 52$ then $k$ pupils from the groups $A$ and $B$ for which $j=78-k$ attend a nature tour.

It can be readily seen that all conditions of the problem are fulfilled.

Now we show that if an $n$ days program satisfies the conditions then $n\le 77$. Let us consider $52$ tours with largest numbers of pupils: $M_1,M_2,\dots ,M_k,N_{k+1},\dots ,N_{52}$ (here each $M_i$ denotes a pupil group attending a museum tour and $N_i$ denotes a pupil group attending a nature tour). By conditions $|M_i\cap N_j|\le 1$ and for $i\ne j$ we have $|M_i\cap M_j|=0$ and $|N_i\cap N_j|=0$. Therefore, $$\sum |M_i|+\sum |N_i|$$ $$=|M_1\cup \cdots \cup M_k\cup N_{k+1}\cup \cdots \cup N_{52}|+\sum |M_i\cap N_j|\le$$ $$2022+k(52-k)\le 2698<26\cdot 104$$ On the other hand, since all tours were attended by different numbers of pupils, we get $$\sum |M_i|+\sum |D_i|\ge n+(n-1)+\cdots +(n-51)=(2n-51)\cdot 26$$ Therefore, $2n-51<104$ and hence $n\le 77$.