Given the natural $n$. We shall call [i]word [/i] sequence from $n$ letters of the alphabet, and [i]distance  [/i] $\rho(A, B)$  between [i]words  [/i] $A=a_1a_2\dots a_n$ and $B=b_1b_2\dots b_n$ , the number of digits in which they differ (that is, the number of such $i$, for which $a_i\ne b_i$). We will say that the [i]word [/i] $C$  [i]lies [/i] between words  $A$ and   $B$ , if $\rho (A,B)=\rho(A,C)+\rho(C,B)$. What is the largest number of [i]words [/i] you can choose so that among any three, there is a [i]word lying[/i] between the other two?
1. **Claim**: The largest number of words you can choose so that among any three, there is a word lying between the other two is \( n + 1 \) for \( n \neq 2 \) and \( n + 2 \) for \( n = 2 \).

2. **Construction for \( n \neq 2 \)**: 
   - Consider all words consisting of \( k \) letters "a" followed by \( n - k \) letters "b" for all \( 0 \leq k \leq n \). This gives us \( n + 1 \) words.

3. **Construction for \( n = 2 \)**:
   - Consider all words consisting of letters "a" and "b". This gives us the words "aa", "ab", "ba", and "bb", which are 4 words in total.

4. **Proof**:
   - We need to show that any set \( S \) of words with the described property has \( |S| \leq n + 1 \) if \( n \neq 2 \) and \( |S| \leq n + 2 \) if \( n = 2 \).

5. **Distance Property**:
   - For any three words \( A, B, C \), we have \( \rho(A, C) + \rho(C, B) \geq \rho(A, B) \) with equality if and only if for each \( i \), either \( c_i = a_i \) or \( c_i = b_i \).

6. **Binary Strings**:
   - For any \( 1 \leq i \leq n \), at most two distinct letters occur in the \( i \)-th position in words in \( S \). Thus, we can assume all words in \( S \) are binary strings of length \( n \).

7. **Case \( n \leq 2 \)**:
   - For \( n = 2 \), the claim follows directly from the construction. So, assume \( n \geq 3 \).

8. **Interpretation as Subsets**:
   - Interpret each string from \( S \) as a subset of the set \( [n] = \{1, 2, \ldots, n\} \). Assume \( S \) contains the string consisting of \( n \) zeros.

9. **Claim 1**: For any distinct non-empty \( A, B \in S \), either \( A \cap B = \emptyset \) or \( A \subset B \), or \( B \subset A \).
   - **Proof**: For any distinct non-empty \( A, B \in S \), either \( \emptyset \) lies between \( A, B \), meaning that \( A, B \) are disjoint, or \( A \) lies between \( \emptyset \) and \( B \), which implies that \( A \subset B \).

10. **Claim 2**: No three non-empty subsets in \( S \) are pairwise disjoint.
    - **Proof**: Assume otherwise, then there exist non-empty \( A, B, C \in S \) such that \( A \cap B = B \cap C = C \cap A = \emptyset \). WLOG \( B \) lies between \( A \) and \( C \), but then taking some \( b \in B \) implies that either \( b \in A \) or \( b \in C \), which is absurd.

11. **Chain Argument**:
    - Let \( S' = S \setminus \{\emptyset\} \). If the subsets in \( S' \) form a chain, we have \( |S'| \leq n \). Suppose this is not the case. Take disjoint \( X, Y \in S' \) so that \( |X| + |Y| \) is maximum. Then for each \( Z \in S' \setminus \{X, Y\} \), either \( Z \subset X \), or \( Z \subset Y \), or \( X, Y \subset Z \).

12. **Bounding \( |S| \)**:
    - If \( [n] \in S' \), then \( X \cup Y = [n] \) and \( S_1, S_2 \) must be empty, thus \( |S'| \leq 3 \leq n \). Suppose \( [n] \not \in S \). The subsets from \( S_1 \) form a chain, so \( |S_1| \leq |X| - 1 \) and similarly \( |S_2| \leq |Y| - 1 \). The subsets in \( S_3 \) form a chain by Claim 1, implying \( |S_3| \leq n - |X| - |Y| \).

13. **Final Calculation**:
    \[
    |S| = |S' \setminus \{X, Y\}| + 3 = |S_1| + |S_2| + |S_3| + 3 \leq (|X| - 1) + (|Y| - 1) + (n - |X| - |Y|) + 3 = n + 1
    \]

The final answer is \( \boxed{ n + 1 } \) for \( n \neq 2 \) and \( n + 2 \) for \( n = 2 \).