$9A_2$, so $H . G = 0$. We can also calculate that $G^2 =
-6$. Now let $L = H + G - \frac{1}{3}\sum_{i=1}^9 (C_i + 2D_i)$. Then $\lceil L \rceil = H + G$ and $\lceil L \rceil^2 = -4$. Clearly, $K_X + L = L \equiv H$ is nef and big. However, by the Kawamata-Viehweg vanishing, and Riemann-Roch theorem, we have $h^0(X, K_X + \lceil L \rceil) = \frac{1}{2} \lceil L \rceil^2 + 2
= 0$.

A similar example can be constructed, if one can find a quartic surface with $16$ nodes (i.e., a normal Kummer quartic surface).

Irregular surfaces
==================

In this section, we shall show that Conjecture \[mainconj\] holds true (with only the first two conditions there but not the last condition) for surfaces $X$ with positive irregularity $q(X)$.

To be precise, let $X$ be a nonsingular projective surface with $q(X)>0$ and let ${\text{\rm alb}}: X \to {\text{\rm Alb}}(X)$ be the Albanese map. Then we have:

\[irreg\] Let $X$ be a nonsingular projective surface with $q(X)>0$. Let $L$ be a nef and big ${\bf Q}$-divisor such that $K_X+L$ is nef. Then $H^0(X,K+ \lceil L \rceil) \ne 0$.

To see this, we need the following lemma:

Let 