TITLE: Smallest diameter of a balanced subset of the Hamming cube
QUESTION [11 upvotes]: Let $\{0,1\}^n$ be the Hamming cube with the Hamming metric. It's a metric space of diameter $n$. 
Let's call a set $B\subset \{0,1\}^n$  balanced if its center of mass is the center of the cube; that is, the average of all vectors contained in $B$ is
$(1/2,\dots,1/2)$. What is the smallest possible diameter of a nonempty balanced subset of $\{0,1\}^n$? 

Partial results
Let $d_n$ be the aforementioned smallest diameter, and let $B$ be a balanced set that realizes it. Without loss of generality, $B$ contains the zero vector $(0,\dots,0)$. To be balanced, it must also contain some vector with more $1$s than $0$s. Hence
$$
d_n \ge \left\lceil \frac{n+1}{2} \right\rceil \tag{1}
$$
The Cartesian product of two balanced sets is balanced, and its diameter is the sum of diameters of its factors. Therefore, the sequence $(d_n)$ is subadditive:
$$
d_{m+n} \le d_m+d_n \tag2
$$
(which in particular implies that $d_n/n$ has a limit.) 
The values of $d_n$ I know so far:
$$
\begin{array}{|c|c|}
\hline 
n & d_n  \\
\hline  
1 & 1 \\
2 & 2 \\ 
3 & 2 \\
4 & 3 \\
5 & 4 \\
6 & 4 \\
\hline
\end{array}
$$
Most of the table is obtained from $(1)$-$(2)$ with the help of the example 
$$
\begin{pmatrix}
0 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
0 & 1 & 1 & 0 
\end{pmatrix}
$$
where the columns form a balanced set of diameter $2$, showing $d_3\le 2$. (By the way, this is a tetrahedron inscribed in a cube.)
For $d_5$, the bounds $(1)$-$(2)$ give $3\le d_5\le 4$. To see that the diameter cannot be $3$, note that WLOG a balanced set of diameter $3$ contains the following columns:
$$
\begin{pmatrix}
0 & 1 \\
0 & 1 \\ 
0 & 1 \\ 
0 & 0 \\
0 & 0 
\end{pmatrix}
$$
Since the total number of $0$s and $1$s in the last two rows must be the same, we need a column ending with two $1$s: 
$$
\begin{pmatrix}
0 & 1 & * \\
0 & 1 & * \\ 
0 & 1 & *\\ 
0 & 0 & 1 \\
0 & 0 & 1
\end{pmatrix}
$$
but no matter how the asterisks are filled, the diameter is greater than $3$.

REPLY [2 votes]: This is the streamlined (incomplete) answer resuming all results in my other answer (and including some results obtained by Michelle). 
I don't think the general problem will be settled in the short term.
First we give improve the lower bound for $d_n\ge \lceil\frac{n+1}{2}\rceil$ given in (1) in the OP:
$\bf{Proposition\ 1:(III)}$ $d_{4k+1}\ge 2k+2$ for $k>0$.
So we have a lower bound that can be described modulo 4:
$n=4k$, $d_n\ge b_{4k}=2k+1$ for $k>0$.
$n=4k+1$, $d_n\ge b_{4k+1}=2k+2$
$n=4k+2$, $d_n\ge b_{4k+2}=2k+2$
$n=4k+3$, $d_n\ge b_{4k+3}=2k+2$
We obtain the following table for the lower bounds:
$$
\begin{array}{|c|c|}
\hline 
n & b_n  \\
\hline  
1 & 1 \\
2 & 2 \\ 
3 & 2 \\
4 & 3 \\
5 & 4 \\
6 & 4 \\
7 & 4 \\
8 & 5 \\ 
9 & 6 \\
10 & 6 \\
11 & 6 \\
12 & 7 \\
13 & 8 \\
14 & 8 \\ 
15 & 8 \\
16 & 9 \\
17 & 10 \\
18 & 10 \\
19 & 10 \\
20 & 11 \\
21 & 12 \\
22 & 12 \\
23 & 12 \\
24 & 13 \\ 
25 & 14 \\
26 & 14 \\
 \vdots&\vdots\\
\hline
\end{array}
$$
$\bf{Proposition\ 2:}$ If for some $j$ we have $d_{4j+3}=b_{4j+3}=2j+2$, then 
$d_n=b_n$ for $n\in\{4j+1,4j+2,4j+4,4j+5,8j+5,8j+6\}$.
The proof follows directly from the lower bounds and (2) in the OP: $d_{n+m}\le d_n+d_m$.
In view of this proposition, in order to establish $d_n=b_n$ for all $n$, it suffices to prove it for $n=4j+3$ for all $j$. 
We will prove the equality only for $4j+3=p$ a prime, $4j+3=p(p+2)$ a product of twin primes or $4j+3=2^t-1$ for some $t$. This proves for example that
$d_n=b_n$ for all $n\le 50$ except $n\in\{27,28,39,40\}$. 
We will describe a construction of balanced sets of vectors of length $4j+3$  of diameter $2j+2$ out of a cyclic $(v,k,\lambda)$ difference set with parameters $(v,k,\lambda)=(4j+3,2j+1,j)$, thus proving that whenever such a difference set exists, we have $d_{4j+3}=b_{4j+3}=2j+2$.
It is known that such cyclic sets exist for the three cases mentioned above (See https://oeis.org/A217332 ).
For the other cases of $n=4j+3$ it is known that no such cyclic difference sets exist for $n\le 10000$, but that doesn't imply that there are no balanced sets of diameter $2j+2$, so these cases remain open.
${\bf{Definition}}$ A cyclic $(v,k,\lambda)$ difference set $D$ is a set of $k$ residues modulo $v$ such that for each non-zero residue $s\mod v$ the equation $x-y\equiv s \mod v$ has exactly $\lambda$ solution pairs $(x,y)$, with $x,y\in D$.
In particular cyclic Hadamard difference sets have the parameters $v=4j+3$, $k=2j+1$ and $\lambda=j$ for a non-negative integer $j$. 
Now define vectors $V_0,\dots,V_{v-1}\in\{0,1\}^v$ by
$$
(V_s)_t=\left\{\begin{array}{cc}
0&\text{ if $t-s\in D$}\\
1&\text{ if $t-s\notin D$}
\end{array}
\right.
$$
${\bf{Claim:}}$ The set $\{\vec{0},V_0,\dots,V_{v-1}\}$ is a balanced set of vectors of diameter $2j+2$.
It is straightforward to check that the set is balanced. Moreover, 
$$
d(\vec{0},V_s)=\#\{t, \ (V_s)_t=1\}=v-\# D=4j+3-( 2j+1)=2j+2,
$$
and $d(V_s,V_i)=d(V_0,V_{|s-i|})$, so it suffices to prove that
$d(V_0,V_s)=2j+2$ for all $s\ne 0$.
Now fix $s$ and note that
$$
d(V_0,V_s)=\#\{t, \ (V_0)_t\ne (V_s)_t\}
$$
$$=\quad\#\{t,\ (V_0)_t=0\text{ and }(V_s)_t=1\}\quad +\quad 
\#\{t,\ (V_0)_t=1\text{ and }(V_s)_t=0\}.
$$
We define 
$$
A_{10}=\{t,\ (V_0)_t=1\text{ and }(V_s)_t=0\}
$$ 
$$
A_{01}=\{t,\ (V_0)_t=0\text{ and }(V_s)_t=1\}
$$ 
$$
A_{00}=\{t,\ (V_0)_t=0\text{ and }(V_s)_t=0\}
$$ 
$$
A_{11}=\{t,\ (V_0)_t=1\text{ and }(V_s)_t=1\}.
$$ 
Then 
$$
\# A_{10}+\# A_{00}=\#D=2j+1\quad\text{and}\quad \# A_{01}+\# A_{00}=\#D=2j+1,
$$
 and so 
$$
\# A_{10}=\# A_{01}=2j+1-\# A_{00}.
$$
But
$$
A_{00}=\{t,\ (V_0)_t=0\text{ and }(V_s)_t=0\}=\{t,\ t\in D\text{ and }t-s\in D\},
$$
and so 
$$
\# A_{00}=\#\{t, \ (t,t-s)\in D^2 \}=\{(x,y)\in D^2,\ x-y=s\}
$$
But, by definition, the equation $x-y\equiv s \mod v$ has exactly $\lambda=j$ solution pairs $(x,y)$, with $x,y\in D$, hence
$\# A_{00}=j$ and so 
$$
d(V_0,V_s)=\# A_{10}+\# A_{01}=2(2j+1-\# A_{00})=2j+2,
$$
as desired.
We can say something about $\lim_{n\to\infty} \frac {d_n}{n}$.
By Fekete's subadditivity lemma and (2) in the OP, the limit $\lim_{n→∞} \frac{d_n}n$ exists. So the fact that there are infinitely many numbers for which the lower bound is attained (e.g., $2^t−1$) implies $\lim_{n→∞} \frac{d_n}n =\frac 12$. 
Finally assume the following variant of the Goldbach conjecture is true:
Every number of the form $4k+2$ is the sum of two primes $p,q$ of the form $4k+3$. 
We know that $d_p=\frac{p+1}2$ and $d_q=\frac{q+1}2$. Hence, if this conjecture is true, the lower bound is attained at every number of the form $4k+2$ and at every number of the form $4k+1$.