This case can be reduced to the one with even clubs and even intersections.
Let's call the maximal number of clubs in the odd/odd and even/even cases $o_n$ and $e_n$, respectively. Gerry has already shown $e_{n-1}\le o_n\le e_{n+1}$. To see that in fact $o_n\le e_n$, consider the clubs as vectors in $\mathbb F_2^n$, indexed by the citizens, pick some club $c$, and form a set $S$ of clubs by adding $c$ to all clubs (including $c$ itself). The resulting vectors are even and have even intersections.
As shown in Bollobas' book, $e_n=2^{\lfloor n/2\rfloor}$. Thus the bounds in $e_{n-1}\le o_n\le e_n$ coincide for odd $n$, so for this case we have $o_n=2^{(n-1)/2}$. For even $n$, note that $S$ does not contain the componentwise complement of any of its elements, since the original clubs corresponding to an element of $S$ and its complement would also be complements of each other, and would thus have even intersection. Thus we can double the size of $S$ by adding the complements of all its elements, and all vectors will still be even and have even intersections (since $n$ is even). Thus in this case $o_n\le e_n/2=2^{n/2-1}$, so that $2^{n/2-1}$ is both a lower and upper bound. In summary, $o_n=2^{\lfloor(n-1)/2\rfloor}$ for either parity of $n$.