Problem 1605

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 $n = 2$.