= Solution
Write $P$ for the <prime numbers> in $A$ and $C=A\setminus P$ for its nonprime elements. Since $n\geq4$, both <antichains> have at least two elements: $2,3\in P$ and $1,4\in C$. Every element of $P$ is below every element of $C$. We can classify a <subset> $S$ by its intersection with these two levels of the <two-level partial order>.
* If $S=\varnothing$, every element of $A$ is an <upper bound>. There is no <maximum>, because $S$ has no elements, and no <least upper bound>, because the distinct elements $2,3$ are incomparable minimal elements of $A$.
* If $S=\{p\}$ with $p\in P$, its <upper bounds> are $\{p\}\cup C$; its <maximum> and <least upper bound> are both $p$.
* If $S\subseteq P$ has at least two elements, its <upper bounds> are exactly $C$. There is no <maximum> in $S$ and no <least upper bound>, because $C$ has at least two incomparable elements.
* If $S\cap C=\{c\}$, its only <upper bound> is $c$: no different element is above this nonprime element. Since every remaining element of $S$ is a <prime number>, $c$ belongs to $S$ and dominates all of it. Thus $c$ is both its <maximum> and its <least upper bound>.
* If $S$ contains at least two elements of $C$, these distinct nonprime elements have no common <upper bound>, and $S$ has neither a <maximum> nor a <least upper bound>.
Consequently \b[an <upper bound> exists exactly when $|S\cap C|\leq1$; a <maximum> or <least upper bound> exists exactly when $|S\cap C|=1$ or $S$ is a singleton <subset> of $P$]. The exhaustive cases also justify why the last two existence criteria coincide here.
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