XXIV OBM

Let $p$, $q$ be two binary sequences. Define $d(p,q)$ as the number of positions $p$ and $q$ differ.

Lemma. $d$ satisfies the triangle inequality, that is, $d(p,q)\le d(p,r)+d(r,q)$ for all binary sequences $p$, $q$, $r$.

Proof. $p$ and $r$ differ in $d(p,r)$ positions, so they coincide in $24-d(p,r)$ positions. Of these positions, in at most $d(q,r)$ positions $q$ and $r$ differ, so $p$, $q$ and $r$ coincide in at least $24-d(p,r)-d(q,r)$ positions. So $d(p,q)\le d(p,r)+d(r,q)$.

Define the sphere with center $p$ as the set of binary sequences $q$ such that $d(p,q)\le 4$. Let $S$ be a set of binary sequences of length $24$ so that any two differ in at least $8$ positions and $p_1$, $p_2$ two binary sequences from $S$. If the spheres with centers $p_1$ and $p_2$ have an intersection, then $p_1$ and $p_2$ are at distance at most $4+4=8$. So $d(p_1,p_2)=8$ and for all sequences $q$ in the intersection $d(p_1,q)=d(p_2,q)=4$. Each sequence $q$ belongs to at most $24/4=6$ spheres with center in $S$, because if $q$ belongs to more spheres then there are two sequences $q_1$ and $q_2$ such that they differ with $q$ in the same positions. This means that $d(q_1,q_2)<d(q_1,q)+d(q,q_2)=8$, contradiction.

The number of sequences in each sphere is $\left(\genfrac{}{}{0ex}{}{24}{0}\right)+\left(\genfrac{}{}{0ex}{}{24}{1}\right)+\left(\genfrac{}{}{0ex}{}{24}{2}\right)+\left(\genfrac{}{}{0ex}{}{24}{3}\right)+\left(\genfrac{}{}{0ex}{}{24}{4}\right)$. Since each pair of spheres can intersect and each sequence belongs to at most $6$ spheres, $$\begin{gathered} |S|\cdot \left(\left(\genfrac{}{}{0ex}{}{24}{0}\right)+\left(\genfrac{}{}{0ex}{}{24}{1}\right)+\left(\genfrac{}{}{0ex}{}{24}{2}\right)+\left(\genfrac{}{}{0ex}{}{24}{3}\right)+\frac{1}{6}\left(\genfrac{}{}{0ex}{}{24}{4}\right)\right)\le 2^{24} \\ ⟺|S|\cdot 4096\le 2^{24}⟺|S|\le 4096 \end{gathered}$$