<s>[INST] Smallest diameter of a balanced subset of the Hamming cube 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 $3$.
 [/INST] (This would probably be better as a comment than an answer but it looks like I can't comment yet.)  
I think I can slightly improve on your partial results, with $d_7=4$ and (hence) $d_8=5$.  (Since whenever $d_{2k-1} = k$ your bounds $(1)$-$(2)$ give $d_{2k}=k+1$.)  
For the first claim, it's enough to find a balanced set with diameter equal to $4$, and it appears that 
$$\left\{ 
\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}  
\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}
\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} 
\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \\ 1 \end{pmatrix} 
\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 1 \end{pmatrix} 
\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \\ 0 \\ 1 \\ 1 \end{pmatrix} 
\begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \\ 1 \\ 0 \\ 1 \end{pmatrix} 
\begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \\ 1 \\ 1 \\ 0 \end{pmatrix}  
\right\}$$
is exactly such a set. 
</s>