Suppose that $B$ and $F$ are connected, simply connected, compact, finite dimensional CW complexes. Assume additionally that both are Poicare duality spaces over $\mathbb{Z}$.
Suppose further that for any odd integer $i$ we have $H^{i}(B,\mathbb{Z}) = H^{i}(F,\mathbb{Z})= 0$.
Let $\pi: E \rightarrow B$ be a (connected) fibre bundle with fibre $F$.
Question: Is it true that $H^{i}(E,\mathbb{Z})= 0$, for any odd integer $i$.
Yes. First, note that $B$ must have even dimension (as a Poincare duality space) and therefore by Poincare duality its homology is also trivial in odd degrees. It then follows by the universal coefficient theorem that the homology groups of $B$ are free (otherwise the torsion would show up as odd-degree cohomology), and so the odd degree cohomology of $B$ vanishes for any choice of coefficients.
Now the conclusion follows immediately from the Serre spectral sequence. There is a spectral sequence whose $E_2$ page is $H^p(B,H^q(F,\mathbb{Z}))$ which converges to $H^{p+q}(E,\mathbb{Z})$. For $H^p(B,H^q(F,\mathbb{Z}))$ to be nonzero, $p$ and $q$ must both be even, in which case $p+q$ is even as well, and so $H^i(E,\mathbb{Z})$ can only be nonzero if $i$ is even. (In fact, more strongly, we can see that all the differentials of the spectral sequence will have either trivial domain or trivial codomain, and so the spectral sequence collapses at $E_2$).