$A, B, C, D, E$ and $F$ are members of a six-member group. The group members share $n$ pieces of chocolate in such a way that each of them gets at least one piece, and $A$ gets fewer than $B$, who gets fewer than $C$, who gets fewer than $D$, who gets fewer than $E$, and finally $F$ gets the most. The group members know these conditions, the value of $n$, and of course how many pieces of chocolate they themselves receive, but they have no other information. What is the smallest value of $n$ for which the distribution can be carried out in such a way that none of them can accurately state how many pieces of chocolate each of the others received?
Solution. Since everyone gets at least one piece of chocolate and $A$ gets fewer than $B$, who gets fewer than $C, \ldots$, who gets fewer than $F$, at least $1+2+3+4+5+6=21$ chocolates were distributed. Let's call this the minimum distribution. This must be included in every distribution, so we need to distribute an additional $n-21$ pieces of chocolate while ensuring the conditions of the problem are met.

When distributing the extra chocolates (those above the minimum), we must consider that if someone gets a certain number of extra chocolates, then everyone following them in the ABC order must get at least that many. We are looking for the number $n$ for which there is no one in the group who can precisely determine how many chocolates the others received.

If $n=22$ chocolates are distributed, then according to the above, only $F$ can get the extra chocolate.

If 23 chocolates are distributed, then two extra chocolates need to be distributed. We will list the possible distributions in a table:

| $E$ | $F$ |
| :---: | :---: |
| 0 | 2 |
| 1 | 1 |

In both cases, $E$ and $F$ can determine the distribution.

If $n=24$: In this case, there are 3 extra chocolates, which can only be received by $D, E$, and $F$. We will list the possible distributions in a table:

| $D$ | $E$ | $F$ |
| :---: | :---: | :---: |
| 1 | 1 | 1 |
| 0 | 1 | 2 |
| 0 | 0 | 3 |

In all three cases, $F$ can determine the distribution.

If $n=25$: In this case, 4 extra chocolates need to be distributed. Now $C$ can also receive some, but $A$ and $B$ still cannot. We will list the possible distributions in a table:

| $C$ | $D$ | $E$ | $F$ |
| :---: | :---: | :---: | :---: |
| 1 | 1 | 1 | 1 |
| 0 | 1 | 1 | 2 |
| 0 | 0 | 1 | 3 |
| 0 | 0 | 2 | 2 |
| 0 | 0 | 0 | 4 |

If $C$ gets an extra chocolate, then $C$ can clearly determine that everyone after them got one, and will state this. If $C$ does not speak, then the others know that $C$ did not get an extra chocolate. In this case, if $D$ gets an extra chocolate, $D$ will speak up, and in all other cases, $F$ can determine the distribution.

This shows that up to $n=25$ chocolates, they can still determine how many chocolates each person has. However, with 26 chocolates, they cannot, for example, in the distribution $1-2-3-4-7-9$, from the perspective of any member, multiple distributions are possible.

Therefore, $n=26$ is the smallest value that meets the conditions of the problem.