This is the question. Is it known that a symplectic $4$fold with $b_2>1$ should have a homology class $C$ with $C^2<0$?

Symplectic geography in 4 dimensions can be mapped using Chern number coordinates $(c_1^2,c_2)$. The part of the plane where $c_1^2 > 4c_2$ is uncharted. It's unknown whether there are any symplectic 4manifolds in this region, besides blowups of ruled surfaces, though by the BogomolovMiyaokaYau inequality and the KodairaEnriques classification, there are no complex surfaces. I can't answer the question but I'll point out that a symplectic 4manifold with $b_=0$ and $b_+>1$ necessarily lies in this unknown region  in particular, it's not Kaehler. To see this, rewrite $c_1^24c_2$ in terms of Euler characteristic $\chi$ and signature $\sigma$ as $(2\chi+3\sigma)4\chi$. For a symplectic manifold with $b_=0$, this quantity equals $4b_1+b_24$ and is positive unless $b_1=0$ and $b_2= 1$ or $3$; I use the parity argument mentioned in Paul's comment. If $b_1=0$ and $b_2=3$ then the intersection form on $H_2/tors.$ is $\mathbb{Z}^3$, the unique rank 3 positivedefinite unimodular lattice. So $c_1^2=15$ is the sum of three squares; but it's not. 

