My question is originally related to coding theory, but fairly easy to state in pure combinatorial way.

Fix $k\in\mathbb{N}$, $\beta\in(0,1)$ and consider the binary cube $\Sigma_n = \{0,1\}^n$ equipped with the Hamming distance. Is it true that there exists nearly equidistant $x_1,\dots,x_k\in\Sigma_n$ with pairwise Hamming distance of $\beta n$. More concretely, is it true that for any $\beta\in(0,1)$ and any $\gamma$ small enough, there is an $N^*$ such that for all $n\ge N^*$ there exists $x_1,\dots,x_k\in\Sigma_n$ such that $$ \bigl|n^{-1} d_H(x_i,x_j)-\beta\bigr|\le \gamma. $$

**Thoughts.** If $\beta \le \frac12$ then probabilistic method takes care of it: assign randomly each coordinate $x_i(k)$ ($1\le k\le n$) of $x_i$ so that $\mathbb{P}[x_i(k)=1]=p$, where $p$ satisfies $2p(1-p)=\beta$. Check that $\mathbb{E}[d_H(\sigma_i,\sigma_j)]=\beta n$. Setting $\mathcal{E}_{ij}$ to be the event that $n^{-1} d_H(x_i,x_j) \in[\beta-\eta,\beta+\eta]$ (which occurs with probability $o_n(1)$) simple union bound over $\binom{k}{2}$ events (which is of constant order in $n$) yields the conclusion for all $n$ large enough. But this argument fails if $\beta>\frac12$ as $\max_{p\in[0,1]} 2p(1-p)=1/2$.

**Follow-up.** Noam's bound is tight for $k$ even. For $k$ odd, we have $s_i = \sum_{1\le j\le k}v_j(i)\equiv 1\pmod{2}$ for each $1\le i\le n$ as $v_j(i)\in\{\pm 1\}$. Namely, the coordinates of sum $s=\sum_j v_j$ are odd, thus $\langle s,s\rangle \ge n$. Hence we get (after sending $\gamma\to 0$)
$$
n\le kn\bigl(1+(k-1)(1-2\beta)\bigr).
$$
Rearranging, we find $\beta\le (k+1)/2k$ for $k$ odd.

**Existence.** Now the existence. Fix coordinate $1\le j\le n$, generate $x_1,\dots,x_k$ randomly according to following distribution: $(x_i(j):1\le i\le k)$, $1\le j\le n$ is i.i.d. with $\mathbb{P}[x_i(j)=1]=1/2$ for all $i,j$ and $\mathbb{P}[x_i(j)=x_t(j)]=1-\beta$ for $1\le i<t\le k$. Now, Iosif Pinelis' answer here shows the existence of such a joint distribution. Under this, it is easily seen $\mathbb{E}[n^{-1}d_H(x_i,x_t)] = \beta$; the rest follows by a simple application of probabilistic method via Chebyshev's inequality.