Answer: $n+1$ for $n \ne 2$ and $4$ for $n = 2$.
For $n \ne 2$, take all words consisting only of letters $a$ and $b$, in which each letter $a$ to the left of each letter $b$, and for $n = 2$ the words $aa, ab, ba, bb$.

Let us prove that more words can not be chosen. Consider two selected words $A$ and $B$, located at the largest distance $m$. Suppose that the selected word $C$ is from $A$ at a distance $k > 0$. Then $\rho(B, C)$ is equal to the sum or difference of $\rho(A, B) = m$ and $\rho(A, C) = k$ and does not exceed $m$, hence $\rho(B, C) = m - k$. Now, if some selected word $C' \ne C$ is also from $A$ at distance $k$, we can assume without loss of generality $k \ge m/2$ (otherwise, we can swap $A$ and $B$). The distance between $C$ and $C'$ can not be equal to $k - k = 0$, therefore, it is equal to $k + k \ge m$, and hence $k = m/2$.

So, we have that for each $k$, $0 \le k \le m$, except, perhaps, for $k = m/2$, among the selected words no more than one, located from $A$ at distance $k$. The selected words, which are at a distance of $m/2$ from $A$, are no more than 2 (we showed that the distance between any two such words is $m$, therefore none of the three words lies between the two others).

It remains to check that if $m > 2$ and two words are chosen at a distance $k = m/2$ from $A$, then there are at most 4 words. Indeed, let $\rho(A, C) = \rho(A, C') = k = m/2$, and we have chosen some other word $D$. As in the previous argument, we can assume without loss of generality that $\rho(A, D) = s < k$. Then $\rho(B, D) = 2k - s$ and $\rho(D, C) = \rho(D, C') = k - s$ (the distance between $D$ and $C$ can not be equal to the sum $\rho(B, D) + \rho(B, C) = 2k - s + k > 2k$, so it is equal to

the difference $\rho(B, D) - \rho(B, C)$). It turns out that among the three distances $k-s, k-s, 2k$, the words $D, C, C'$ are not equal to the sum of the other two, a contradiction.
This proves the estimate: if $m > 2$, then for each $k \le n$ there is at most one selected word at a distance of $k$ from $A$, and there are no more than $n+1$ words in total, and if $m = 1$ or $2$, then there are no more than 4 words.