When does an even-dimensional manifold fiber over an odd-dimensional manifold? 
Are there simple necessary and sufficient conditions for an (oriented) even-dimensional compact smooth manifold to fiber over an (oriented) odd-dimensional manifold (with oriented fibers)?

For example, if $M \to N$ is a fiber bundle of compact manifolds with fiber $F$, then their Euler characteristics satisfy $\chi(M) = \chi(N)\chi(F)$.  But if $N$ and $F$ are odd-dimensional, $\chi(N) = \chi(F) = 0$, and so $\chi(M)$ must also vanish.  Is the converse true?  I.e. if $M$ has vanishing Euler class, does it fiber over an odd-dimensional manifold?  Or perhaps there's a hint that $\chi(M) = 0^2$, not just $0$, and so maybe some combination of Massey products also must vanish?
 A: No, I don't believe there's a simple solution. But
here's an approach to the problem which indicates how it can be fractured up.
Assume $M,N$ are closed and connected. 
If $f\: M \to N$ is homotopic to a smooth fiber bundle
with $M$ and $N$ compact, then the fibers are homotopy finite (i.e., they are homotopy equivalent to a finite complex). 
Conversely, it is a result first stated by Quinn (later proved by Gottlieb, and then differently by me) that if $f\: M^m \to N^n$ is such that its homotopy fiber $F$ (at some basepoint in $N$) is homotopy finite, then $F$ is a Poincare duality space of dimension $m-n$. Thus, $f$ gives rise to a fibered surgery problem. 
One can approach this problem in two steps:
Step 1: find a block bundle  $E \to N$ and a fiber homotopy equivalence $E\simeq M$.
This step can be attacked classical surgery techniques (here the dimension of the fiber should be $\ge 6$).  What one studies here is the map $\tilde S_N(M) \to \tilde S(M)$ from the fiberwise block structure space to the block structure space. 
Step 2: Study the map $S_N(M) \to \tilde S_N(M)$ from the fiberwise structure space to the fiberwise block structure space. This step involves higher algebraic $K$-theory a la Waldhausen. This step is only really understood in the "concordance stable range" which in this case requires $4n \le m$ (approximately).${}^\dagger$
The above is only meant to be an outline. I first learned about these ideas from the papers of Weiss and Williams, most notably:
Automorphisms of manifolds. Surveys on surgery theory, Vol. 2, 165–220, Ann. of Math. Stud., 149, Princeton Univ. Press, Princeton, NJ, 2001
An alternative approach which packages Step 1 and Step 2 into a single step
is in the third WW paper which can be obtained from Michael Weiss' website.
More recently, see the papers of Wolfgang Steimle, especially
Obstructions to stably fibering manifolds.
Geom. Topol. 16 (2012), no. 3, 1691–1724
${}^\dagger$ Added Later: According to Steimle, the "stable range" for the fibering problem is more complicated than what I wrote above. Rather than write it down, let me refer to his paper for the actual range.
A: Being the total space of a fiber bundle is not invariant under homotopy equivalences, so I doubt there is a criterion of the type you state (eg in terms of homological invariants).
However, one can say something a little weaker.  If a manifold $M^k$ fibers over a manifold $N^{\ell}$, then the fibers form a codimension $\ell$ foliation of $M^k$.  A deep theorem of Thurston says that if $\chi(M^k)=0$, then $M^k$ supports a codimension $1$ foliation (the converse is an easy exercise).  By the way, to appreciate how deep this is, it implies in particular that an odd-dimensional closed oriented manifold always supports a codimension $1$ foliation.  This is nontrivial to prove even in dimension $3$ (where it was first proved by Lickorish).
