Team Selection Test for IMO 2023

a. We show that $n=2023$ is impossible. Let $\mathcal{S}=\{S_1,S_2,\dots ,S_{2023}\}$ and $T=\{1,2,\dots ,2023\}$. We call a non-empty set $I\subseteq T$ as nice if every non-empty subset $J\subseteq I$ satisfies ${\Delta }_{j\in J}{S}_j\ne \varnothing$. Note that any set of size one is nice, let $L$ be the largest nice subset of $T$. Due to the condition given in the problem, we find ${\Delta }_{j\in J}{S}_j\in \mathcal{S}$ for all non-empty subsets $J\subseteq L$. On the other hand, for non-empty subsets $J_1,J_2\subseteq L$, $J_1\ne J_2$, we have $J_1\Delta J_2\ne \varnothing$, which implies that $$({\Delta }_{j\in J_1}{S}_j)\Delta ({\Delta }_{j\in J_2}{S}_j)={\Delta }_{j\in J_1\Delta J_2}{S}_j.$$ As $J_1\Delta J_2\subseteq L$, we get ${\Delta }_{j\in J_1\Delta J_2}{S}_j\ne \varnothing$. Therefore, for every non-empty $J\subseteq L$, ${\Delta }_{j\in J}{S}_j$ is a distinct element of $\mathcal{S}$. Now, given $i\in T$, if $S_i\ne {\Delta }_{j\in J}{S}_j$ for every non-empty $J\subseteq L$, then we would have $i\notin L$ and $L\cup \{i\}$ is also nice, which is impossible. Hence, $\mathcal{S}=\{{\Delta }_{j\in J}{S}_j:J\subseteq L,J\ne \varnothing \}$, which implies that $2023=2^{|L|}-1$, a contradiction.

b. We show that $n=1023$ can happen, and the answer is all positive integers $k$ satisfying $512\mid k$. Let us take $A\in \mathcal{S}$ and $x\in A$. Define $X\subseteq \mathcal{S}$ as $x\in X\iff X\in X$ and $\mathcal{S}\setminus X=Y$. For every $X\in X\setminus \{A\}$, we have $x\notin A\Delta X$, so we get $A\Delta X\in Y$. Similarly, for every $Y\in Y$, we have $x\in A\Delta Y$, so we get $A\Delta Y\in X\setminus \{A\}$. As a result, we find $|X\setminus \{A\}|=|Y|$, which means $|X|=512$ as $|\mathcal{S}|=1023$. In other words, any $x\in {\bigcup }_{S\in \mathcal{S}}S$ lies on exactly 512 sets. Letting $N={\bigcup }_{S\in \mathcal{S}}S$, we find $1023k=512N$, which gives $512\mid k$.

Let $k=512m$ for some $m\in \mathbb{N}$, and let $L=\{1,2,\dots ,10\}$. Start with empty 10 sets $\{S_1,S_2,\dots ,S_{10}\}$, and then for every non-empty $R\subseteq L$, we put $m$ different elements in $$\left(\bigcap _{r\in R}{S}_r\right)\setminus \left(\bigcup _{r\in L\setminus R}{S}_r\right).$$ Now, for every non-empty $J\subseteq \{1,2,\dots ,10\}$, we have $${\Delta }_{j\in J}{S}_j=\bigcup _{R\subseteq L,|R\cap J|\text{ odd}}{L}_R.$$ Moreover, there are exactly 512 subsets $R\subseteq L$ satisfying $|R\cap J|$ odd, which means that $|{\Delta }_{j\in J}{S}_j|=512m$. On the other hand, for non-empty $J_1\ne J_2$, we have $$({\Delta }_{j\in J_1}{S}_j)\Delta ({\Delta }_{j\in J_2}{S}_j)={\Delta }_{j\in J_1\Delta J_2}{S}_j$$ as $J_1\Delta J_2\ne \varnothing$. Since $|{\Delta }_{j\in J_1\Delta J_2}{S}_j|=512m$, it is non-empty, which shows that ${\Delta }_{j\in J_1\Delta J_2}{S}_j\in \mathcal{S}$. As a result, the set $$\mathcal{S}=\{{\Delta }_{j\in J}{S}_j:J\subseteq L,J\ne \varnothing \}$$ contains exactly 1023 sets of size $512m$ which satisfies the desired condition.