, 2023

a. We show that $n = 2023$ is impossible. Let $\mathcal{S} = \{S_1, S_2, \ldots, S_{2023}\}$ and $T = \{1, 2, \ldots, 2023\}$. We call a non-empty set $I \subseteq T$ as nice if every non-empty subset $J \subseteq I$ satisfies $\Delta_{j \in J} S_j \neq \emptyset$. 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 $J_1, J_2 \subseteq L$, $J_1 \neq J_2$, we have $J_1 \Delta J_2 \neq \emptyset$, 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 $\Delta_{j \in J_1 \Delta J_2} S_j \neq \emptyset$. 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 $S_i \neq \Delta_{j \in J} S_j$ 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, $\mathcal{S} = \{\Delta_{j \in J} S_j : J \subseteq L, J \neq \emptyset\}$, 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 $x \in X \Leftrightarrow X \in X$ 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 $L = \{1, 2, \ldots, 10\}$. Start with empty 10 sets $\{S_1, S_2, \ldots, S_{10}\}$, 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 $J \subseteq \{1, 2, \ldots, 10\}$, 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 $J_1 \neq J_2$, 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 $J_1 \Delta J_2 \neq \emptyset$. 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
$$\mathcal{S} = \{\Delta_{j \in J} S_j : J \subseteq L, J \neq \emptyset\}$$
contains exactly 1023 sets of size $512m$ which satisfies the desired condition.