Even, non liftable Stiefel-Whitney class 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 $c_i\in H^{2i}(M,\mathbb Z)$ .
 A: $\newcommand{\Z}{\mathbb Z}$Let $X = K(\Z/4,2)\times_{K(\Z/2,2)} BSO(3)$ be the classifying space for $3$-dimensional oriented vector bundles together with a lift of the second Stiefel-Whitney class to $H^2(-;\Z/4)$; in particular, there is a canonical vector bundle $E\to X$ classified by $X\to BSO(3)$. By design, there is a fiber sequence $BSU(2)\to X\to K(\Z/4,2)$, and the Serre spectral sequence shows that the map $H^*(K(\Z/4,2);A)\to H^*(X;A)$ is an isomorphism for $*\le 2$. In particular, $H^2(X;A) \cong\operatorname{Hom}(\Z/4,A)$; for $A = \Z$, this vanishes, while for $A = \Z/2$ the map $H^2(BSO(3);\Z/2)\to H^2(X;\Z/2)$ is an isomorphism, so that $w_2(E)\neq 0$, which implies that $w_2(E)$ does not lift to integral cohomology. Note also that the reduction map $H^2(X;\Z/4)\to H^2(X;\Z/2)$ is surjective.
The first Stiefel-Whitney class $w_1(E)$ vanishes by design, and by the Wu formula the third Stiefel-Whitney class is $w_3(E) = \operatorname{Sq^1} w_2(E)$. The cohomology operation $\operatorname{Sq}^1$ is the Bockstein of the short exact sequence $0\to \Z/2\to\Z/4\to\Z/2$; in particular, it vanishes on a class $x$ iff $x$ lifts to $H^*(-;\Z/4)$. By construction, this gives $w_3(E) = \operatorname{Sq}^1(w_2(E)) = 0$, so that all odd Steenrod classes vanish.
In your question, you asked for an even-dimensional vector bundle over a manifold; for this, take the $d$-dimensional part of the skeleton of a CW model of $X$ for a large ($d = 4$ should be enough) finite number $d$, embed it into a large Euclidean space and take a regular neighbourhood such that $X$ is a strong deformation retract, then add a trivial line bundle to (the pullback of) $E$.
In general, you are asking whether the restriction of $w_{2i}$ to the fiber of the odd Stiefel-Whitney classes $BSO(2n)\to \prod_{j=1}^n K(\Z/2,2j-1)$ lifts to $K(\Z,2j)$. This question can be attacked with obstruction theory, which leads directly to the above counterexample.
A: 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.)
