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
$$A = \{(i, j) : i, j \in \mathbb{Z}_+, i + j \leq 52\}$$
To each pupil from $B$ we assign a different pair from the set
$$B = \{(i, j) : i, j \in \mathbb{Z}_+, i \leq 26, j \leq 26\}.$$
Now for each $1 \leq k \leq 77$ in the $k$-th day:

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

†† if $k \geq 52$ 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 $n \leq 77$. 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 $|M_i \cap N_j| \leq 1$ and for $i \neq j$ we have $|M_i \cap M_j| = 0$ and $|N_i \cap N_j| = 0$. Therefore,
$$\sum |M_i| + \sum |N_i|$$
$$= |M_1 \cup \dots \cup M_k \cup N_{k+1} \cup \dots \cup N_{52}| + \sum |M_i \cap N_j| \leq$$
$$2022 + k(52 - k) \leq 2698 < 26 \cdot 104$$
On the other hand, since all tours were attended by different numbers of pupils, we get
$$\sum |M_i| + \sum |D_i| \geq n + (n-1) + \dots + (n-51) = (2n-51) \cdot 26$$
Therefore, $2n-51 < 104$ and hence $n \leq 77$.