A candidate key must be minimal.
Therefore, if one candidate key is a subset of another set of attributes, the larger set cannot also be a candidate key.
To maximize the number of candidate keys, we therefore want the largest possible collection of subsets such that no subset contains another.
For $n$ attributes, the maximum occurs at the middle level of the subset lattice.
Hence, the maximum number is ${}^{n}C_{\lfloor n/2 \rfloor}$
For $n=6$,
${}^{6}C_{3} = \dfrac{6!}{3!3!} =\dfrac{6\times5\times4}{3\times2\times1}=20$
Thus, all $3$-attribute subsets can simultaneously form the largest possible collection of candidate keys.
$\therefore$ Answer $:\boxed{20}$
