Thank you.

The following gives an answer in the case that all the curves are assume to have genus $g=1$, and $b\geq 2$.

One of the curves, say $C=C_1$ has $C^2>0$. As $p_g=q=0$ we have $\chi(O_X)=1-q+p_g=1$. By Riemann-Roch, we have $\chi(C)=1+ \frac12 (C^2-K\cdot C)$. As $C^2+K\cdot C=2g(C)-2=0$ (by adjunction), we have $\chi(C)=1+C^2$. Also as $h^0(K)=p_g=0$ and $C$ is effective, we have $h^2(C)=h^0(K-C)=0$. From the exact sequence $0\to O_X\to O_X(C) \to O_C(C)\to 0$, taking the long exact sequence in cohomology, we have $H^1(X,O(C))=H^1(C, O_C(C))$. If $C^2>0$ then the last group vanishes, hence $h^1(C)=0$. So $\chi(C)=h^0(C)=1+C^2>1$, and there is a pencil of curves of genus $1$. If necessary, blow up at points of $C$ to arrange $C^2=1$. The bundle $O_C(C)$ is of the form $O_C(p_0)$, for some point $p_0\in C$.
The pencil defined by the linear series $|C|$ is a pencil of genus $1$ curves passing through $p_0$. Blowing up at this point we get a surface $X'$ and a map $\pi:X'\to P^1$. Let $\sigma$ be the exceptional divisor, which is a rational curve of self-intersection $-1$. This gives the structure of an elliptic surface, possibly non-minimal, and containing all our genus $1$ curves. Blow down any $(-1)$-curve in a fiber of $\pi$ to get a minimal elliptic surface $X''\to P^1$. This has $q=0, p_g=0$ and it has a section (the image of $\sigma$). By the Enriques-Kodaira classification, this has to be a rational surface. Any section of a minimal elliptic surface has to have self-intersection $-1$ (let $s$ be a section; the canonical bundle is $K=-f$; then $s^2+K\cdot s=-2$, so $s^2=-1$). So $\sigma$ cannot intersect any rational curve being blown-down when going from $X'$ to $X''$, since otherwise its self-intersection increases. All the genus one surfaces $C_i\subset X$ give genus one surfaces $C_i'\subset X'$ which map down to $C_i''\subset X''$. These must be fibers (any curve not intersecting $C$ is contained in a fiber, so it is a rational curve or a fiber). This means that $C_i''$ intersect $\sigma$,
and hence $C_i'$ intersects $\sigma$. When going to $X$, $C_i$ intersects $C$. Contradiction!

simply connected4-manifolds. However, there are (non rational) complex surfaces with $p_g=q=0$ that arenotsimply connected (for instance, the classical Godeaux surface), so in this case I do not see how to conclude that the intersection form must be necessarily diagonalisable. I'm missing something? $\endgroup$ – Francesco Polizzi Mar 23 '15 at 8:51rational(co)homology. A quadratic form over $\Bbb{Q}$ is always diagonalizable. $\endgroup$ – abx Mar 23 '15 at 14:22