I think you can get as many as you like.  Let X be 0 or 1 with probability 1/2 each.  Let $Z|X=I$ be of the form $h \pm \epsilon q$ where q is arbitrary and can be positive or negative, and I write them as if they have a density but any measure/signed measure will, subject to the sum being a measure.  First I claim that 
$$P(X=1| Y=k ) = \frac 12 \frac{\int x^k(1-x)^{N-k}(h(x)  +  \epsilon q(x)) dx)}{\int x^k(1-x)^{N-k}(h(x) dx}$$


$$ = \frac 12 +  \frac{\int x^k(1-x)^{N-k}  \epsilon q(x) dx)}{\int x^k(1-x)^{N-k}(h(x) dx}$$

Second, any f(k) can be written $\int x^k(1-x)^{N-k}   q(x) dx$ for some measure q.  Because, taking q to be a pointmass and you get a term like $\lambda^k$, and most choices of $\lambda_1,..., \lambda_N$ will give you a basis for sequences of length N.  Choose a q that makes it 1,-1,1, -1 etc.  Choose $\epsilon$ really small, so that $h \pm \epsilon q$ is a measure, but h is still more or less arbitrary.  Choose h so that the denominator varies much slower than the numerator, and you ought to  get $P(x=1|y=k)$ being alternately a little bigger that 1/2 and a little less than 1/2.

Proposer asks: can you be more explicit etc.
Let me illustrate in the simplest case, N=1.  I'm going to put mass a at 1/2 and b  3/4.  Then I have 2 equations in 2 unknowns, which  come from specializing  $$ f(k) =  \int p^k(1-p)^{N-k} dq(p) $$ in the case N=1 to $$ f(0) = -1 = \frac 12 a + \frac 14 b $$ and $$ f(1) = 1 = \frac 12 a + \frac 34 b$$.  It has a solution, which is a $\it{signed}$ measure.
In general you'd just want to observe that the system of equations has a solution.