TITLE: Even, non liftable Stiefel-Whitney class
QUESTION [7 upvotes]: Let $M$ be a smooth manifold and $E$ a smooth real vector bundle of even rank over $M$.
If $E$ admits of a complex vector bundle structure $\mathcal E$ ($\mathcal E_\mathbb R=E$) then all odd Stiefel-Whitney classes of $E$ vanish: $$w_{2i+1}(E)=0$$
Moreover the even Stiefel-Whitney classes of $E$ are the images under the reduction morphism $\operatorname {red}^{2i}:H^{2i}(M,\mathbb Z)\to H^{2i}(M,\mathbb F_2)$ of its Chern classes, namely $$w_{2i}(E)=\operatorname {red}^{2i}(c_i(\mathcal E))$$
My question
Is there a real vector bundle of even rank $E$ with all odd $w_{2i+1}(E)=0$ that nevertheless cannot be endowed with a complex structure just because some $w_{2i}(E)\in H^{2i}(M,\mathbb F_2)$ cannot be lifted to $\mathbb Z$?
Explicitly, the equation $$\operatorname {red}^{2i}(c_i)=w_{2i}(E)\in H^{2i}(M,\mathbb F_2)$$ has no solution $c_i\in H^{2i}(M,\mathbb Z)$ .

REPLY [3 votes]: As a complement to Bertram Arnold's excellent answer, one could add a reference to
Teichner, Peter, 6-dimensional manifolds without totally algebraic homology, Proc. Am. Math. Soc. 123, No. 9, 2909-2914 (1995). ZBL0858.57033.
In particular, Lemma 2 states that if $M$ is a closed $4$-manifold with $\pi_1(M)=\mathbb{Z}/4$, then there exists a rank $3$ vector bundle $E$ over $M$ with $w_1(E)=w_1(M)$ and $w_3(E)=0$, and $w_2(E)$ is not the reduction of a class in $H^2(M;\mathbb{Z}^{w_1(E)})$ (where $\mathbb{Z}^{w_1(E)}$ denotes the local system of integer coefficients twisted by $w_1(E)$). Thus over any orientable $4$-manifold with $\pi_1=\mathbb{Z}/4$ we get an example. (Take direct sum with a trivial line bundle to get an even rank example, as in Bertram's answer.)