Find binary number with max hamming distance wrt given set of binary numbers
Here is the decision version of the problem, which we can the problem of farthest string.


  Given $m$ length-$n$ binary strings $s_1, s_2, \cdots, s_m$ and a number $k$, determine whether there is a length-$n$ string $s$ such that $d(s,s_i)\ge k$ for all $i$.


There is the problem of closest string.


  Given $m$ length-$n$ binary strings $s_1, s_2, \cdots, s_m$ and a number $k$, determine whether there is a length-$n$ string $s$ such that $d(s,s_i)\le k$ for all $i$.


Farthest string is dual to closest string.

Let $c$ denote the map that changes a binary string to its 1's complement, i.e., switching 0 and 1 everywhere. For example, $c(010010111)=101101000$. Note that $c\circ c = Id$.

Given an instance of farthest string $(m, n, s_1, s_2, \cdots, s_m, k, s)$, we can construct an instance of closest string $(m, n, c(s_1), c(s_2), \cdots, c(s_m), n-k, s)$. Call this transformation of instances $r$.

For every string $s$ of length $n$, $$d(s,s_i)+d(s, c(s_i))=n.$$
Hence 
$$d(s,s_i)\ge k\text{ for all }i\  \Longleftrightarrow\ d(s,c(s_i))\le n-k\text{ for all }i. $$

So $r$ converts a yes-instance of farthest string to a yes-instance of closest string. Symmetrically, $r$ converts a yes-instance of closest string to a yes-instance of furthest string. Furthermore, $r\circ r= Id$.

So $r$ is a linear duality reduction between farthest string and closest string.

All results about closest string are about farthest string.

That is because of the duality reduction. Now we can dive into the Wikipedia article on closest string to see find what we have about closest string. 

In particular, we know that farthest string is NP-complete. We can check approximability, fixed-parameter tractability as well.