Solution:

Let $C$ be the set of clubs; each student then corresponds to a subset of $C$ (the clubs to which that student belongs). The two-student condition implies that these subsets must be all distinct. Now (assuming there is more than one student) some student belongs to a nonempty set $S$ of clubs. For every subset $T \subseteq C$, let $f(T)$ be the subset of $C$ consisting of those clubs that are in exactly one of $S$ and $T$ (so that $f(T) = (S \cup T) - (S \cap T)$). It is straightforward to check that $f(f(T)) = T$ and $f(T) \neq T$, so that the collection of all $2^{10}$ subsets of $C$ is partitioned into pairs $\{T, f(T)\}$. Moreover,
as long as $S$ is distinct from $T$ and $f(T)$, every club is in either none or exactly two of the sets $S, T$, and $f(T)$, so we cannot have a student corresponding to $T$ and another corresponding to $f(T)$. This puts an upper bound of 513 possible students (one for $S$, one for $\emptyset = f(S)$, and one for each of the 511 other pairs). On the other hand, if we take some club $c$, we can have one student belonging to no clubs and 512 other students all belonging to $c$ and to the 512 possible subsets of the other 9 clubs, respectively. It is readily checked that this arrangement meets the conditions - for the three-student condition, either all three students are in $c$, or one is the student who belongs to no clubs and we reduce to the two-student condition - so 513 is achievable.