There are 25 masks of different colours. k sages play the following game. They are shown all the masks. Then the sages agree on their strategy. After that the masks are put on them so that each sage sees the masks on the others but can not see who wears each mask and does not see his own mask. No communication is allowed. Then each of them simultaneously names one colour trying to guess the colour of his mask. Find the minimum k  for which the sages can agree so that at least one of them surely guesses the colour of his mask.
( S. Berlov )

To solve this problem, we need to determine the minimum number of sages, \( k \), such that at least one of them can always guess the color of their mask correctly, given the constraints.

1. **Understanding the Problem:**
   - There are 25 masks of different colors.
   - Each sage can see the masks on the others but not their own.
   - The sages must guess the color of their own mask without communication after the masks are put on.

2. **Formulating the Strategy:**
   - Let \( C = \{ c_1, c_2, \ldots, c_{25} \} \) be the set of colors.
   - Each sage needs to guess their own mask color based on the colors they see on the other \( k-1 \) sages.

3. **Defining the Functions:**
   - Let \( S \) be the set of all \((k-1)\)-subsets of \( C \).
   - We need to find \( k \) functions \( f_i: S \rightarrow C \) for \( i = 1, 2, \ldots, k \) such that:
     \[
     \forall D = \{ d_1, d_2, \ldots, d_k \} \subset C, \exists i \in \{ 1, 2, \ldots, k \} \text{ such that } f_i(D - \{ d_j \}) = d_j.
     \]

4. **Counting the Possibilities:**
   - There are \( k \times \binom{25}{k} \) possible choices for \( d_j \) and \( D \) in condition (A).
   - Each function \( f_i \) must cover at least \( \binom{25}{k} \) members in its domain.

5. **Inequality Analysis:**
   - We need \( \binom{25}{k-1} \geq \binom{25}{k} \).
   - Using the binomial coefficient property:
     \[
     \binom{25}{k-1} = \frac{25!}{(k-1)!(25-(k-1))!} = \frac{25!}{(k-1)!(26-k)!}
     \]
     \[
     \binom{25}{k} = \frac{25!}{k!(25-k)!}
     \]
   - The inequality becomes:
     \[
     \frac{25!}{(k-1)!(26-k)!} \geq \frac{25!}{k!(25-k)!}
     \]
     \[
     \frac{1}{(k-1)(26-k)} \geq \frac{1}{k}
     \]
     \[
     k \geq 13
     \]

6. **Conclusion:**
   - The minimum \( k \) for which the sages can agree on a strategy such that at least one of them surely guesses the color of their mask is \( k = 13 \).

The final answer is \( \boxed{13} \).