Given a positive integer $n$, the [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) $d^H\_n(x,y)$ of $x,y\in \{0,1\}^n$ is defined by $$d^H\_n(x,y) = |\{k\in\{0,\ldots,n-1\}: x(k)\neq y(k)\}|.$$ 
We say that a positive integer $s$ is $n$-*spreadable* if there is $T\subseteq \{0,1\}^n$ with $|T|=n$ and for $x\neq y\in T$ we have $d\_H(x,y) \geq s$. For any integer $n\geq 1$ let $m\_n$ be the largest $n$-spreadable number less or equal to $n$.


**Question.** Do we have $\lim \sup\_{n\to\infty}\frac{m\_n}{n} = 1$?



Spreading $n$ points in $\{0,1\}^n$ as far as possible
If I understood the question correctly, what you're asking is related to the maximum distance of binary codes with large minimum distance $d\geq s$.


In coding theory, $A\_q(n,d)$ is defined as the maximum cardinality of a $q-$ary code with length $n$ and minimum distance $d.$


You can never have more than 2 codewords at full distance by binary geometry. In fact,
the Plotkin bound for high distance binary codes states:


If $d$ is even (thus $n$ even for your case) and $2d>n\geq d,$ then
$$
A\_2(n,d)\leq 2\left\lfloor \frac{d}{2d-n}\right\rfloor,
$$
which will give $A\_2(n,d)\leq 2,$ for your case. Take any vector and it's bitwise complement.


The $d$ odd case is similar, see for Example Roman's book on Coding and Information Theory.


Of course you only want a lim sup tending to 1, and **you can get it to tend to $1/2$** by using the rows of Hadamard matrices for $n$ a power of 2, but I doubt that any value larger than $1/2$ in your expression is achievable (see update below).


**Edit:** Thanks to Aaron Meyerowitz for clarifying the finite odd length case.


**Proposition:** The lim sup is actually $1/2.$


Assume that a value $d$ larger than $n/2$ is achievable. Map the codewords to $\pm 1$ vectors by writing $((-1)^{x\_1},\ldots,(-1)^{x\_n}).$ The inner product pf two $\pm 1$ valued vectors at hamming distance $d$ from each other is $\delta=n-2d.$ Therefore, if a collection of $m$ distinct $\pm 1$ vectors have minimum distance $d,$ we can write 
$$
0\leq \left| \sum\_{i=1}^m u\_i \right|^2 = \langle \sum\_i u\_i , \sum\_i u\_i\rangle = \sum\_i |u\_i|^2 + 2 \sum\_{i<j} \langle u\_i,u\_j \rangle \leq mn + m(m-1) (n-2d),
$$
which eventually yields
$$
d\leq \frac{n}{2}\frac{m}{(m-1)}. 
$$
Letting $m=n$ proves the claim.


