Suppose $a_1, a_2, \ldots, a_m$ are different strings of the same length n.
And let $V = [v_1, v_2, \ldots, v_n]$ be a matrix such that $V_{i, j}$ is a Hamming distance between $a_i$ and $a_j$.
Hypothesis is that $v_1$ can't be equal to $\alpha_2 v_2 + \ldots + \alpha_n v_n$, where $\alpha_2 + \ldots + \alpha_m = 1$.
If this hypothesis is correct or even incorrect one more complex problem will be solved.
Do you have any ideas about it?
It's getting late here, so I may have missed something. But I think the following example answers the question in the negative.
Let the alphabet be $\{a,b\}$ and consider the set of all strings of length $2$: $a_1=aa$, $a_2=ab$, $a_3=ba$, $a_4=bb$. Then we get $$ V=\left(\begin{array}{cccc} 0&1&1&2\\ 1&0&2&1\\ 1&2&0&1\\ 2&1&1&0\end{array}\right). $$ Here we have the linear dependency relation among the columns $$ v_1=v_2+v_3-v_4, $$ which is of the undesired type.