Do you want to construct an even potential V(x) such that ψ(x=0) = 0?
In an attractive potential having bound states there are two reasons for ψ(x) = 0; one is the generalization of standing waves like sin(kx) or cos(kx); of course you will never get an even wave function for such a standing wave b/c its Taylor expansion would start with the x1 term.
The second possibility is that the potential forces the wave function to zero; but b/c in QM you can have nonzero ψ even in classically forbidden regions (think about a double-well potential constructed like V(x) ~ (x+a)(x-a)) the only way to force the wave function ψ(x) to zero is a divergent potential V(x) → ∞.
So one simple possibility would be a combination of to infinite square wells with V=0 for x in [-b,-a] and x in [+a,+b], and V=∞ outside these two wells.