TITLE: Is there a closed non-smoothable 4-manifold with zero Euler characteristic?
QUESTION [10 upvotes]: I will just repeat the title:

Is there a closed non-smoothable 4-manifold with zero Euler
  characteristic?

I am guessing yes simply based on other existence theorems I have seen for 4-manifolds.

REPLY [12 votes]: Alternatively, one can start with the $E_8$-manifold and connect sum with (five copies of) $S^1\times S^3$, and appeal to Donaldson's diagonalisation theorem instead.
More precisely, the (negative) $E_8$-plumbing $P$ bounds the Poincaré homology sphere $Y$; by Freedman's theorem, $Y$ is also the boundary of a contractible topological 4-manifold $W$, and gluing $P\cup_Y -W$ yields a closed, simply connected topological 4-manifold $X$ with intersection form $-E_8$; $\chi(X) = 10$.
The connected sum $X\#5(S^1\times S^3)$ has Euler characteristic 0, by Mike's computation, and the intersection form is still $-E_8$. Hence, by Donaldson's theorem, $X\#5(S^1\times S^3)$ does not have a smooth structure, since its intersection form is negative definite but not diagonal.

REPLY [11 votes]: You can also get this from Donaldson's theorem by a similar device. Take a non-diagonalizable definite form with even rank $2n$, and realize it (Freedman again) by a simply connected manifold. Then $W\ \#\ (n+1) (S^1 \times S^3)$ is not smoothable and has Euler characteristic $0$. The argument is that if it were smoothable, then you could surger away the fundamental group and realize that intersection form by a simply connected manifold.