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I think that there cannot exists a surface like that. Probably this

EDIT: The proof below is clear and written somewherewrong, but I feel like trying to give a proof:because it is false that $h^0(X',-mK_X')=h^0(\mathbb{P}^2,-mK_{\mathbb{P}^2})$ (see comments)

Consider a normal rational surface $X$ with Gorenstein (or $\mathbb{Q}$-Gorenstein) singularities. Let $\mu:X'\to X$ be a resolution. Then $\mu^*(K_X)=K_{X'}+E$, where $E$ is a $\mu$-exceptioanl divisor. If $X$ is very singular it can happen that $E$ is effective, but it is an exceptional divisor, so that it is not big, or, in other words, it cannot be in the interior of the pseudoeffective cone, or, in other words, given any $A$ is an ample ($\mathbb{Q}$-)divisor, for sure $E-A$ is not effective (it is not pseudoeffective in fact).

On the other hand, as $X'$ is a smooth rational surfaces it should be easy to see that, for all $m\in \mathbb{N}$, $h^0(X', -mK_X')=h^0(\mathbb{P}^2,-mK_{\mathbb{P}^2})$, so that $-K_{X'}$ is in fact big, that is $-K_{X'}\geq H$, for some ample $H$.

Hence $\mu^*(K_X)=K_{X'}+E\leq E-H$ is not pseudoeffective, so that it cannot be effective, and the same holds for $K_X$.

Does it make sense?

1

I think that there cannot exists a surface like that. Probably this is clear and written somewhere, but I feel like trying to give a proof:

Consider a normal rational surface $X$ with Gorenstein (or $\mathbb{Q}$-Gorenstein) singularities. Let $\mu:X'\to X$ be a resolution. Then $\mu^*(K_X)=K_{X'}+E$, where $E$ is a $\mu$-exceptioanl divisor. If $X$ is very singular it can happen that $E$ is effective, but it is an exceptional divisor, so that it is not big, or, in other words, it cannot be in the interior of the pseudoeffective cone, or, in other words, given any $A$ is an ample ($\mathbb{Q}$-)divisor, for sure $E-A$ is not effective (it is not pseudoeffective in fact).

On the other hand, as $X'$ is a smooth rational surfaces it should be easy to see that, for all $m\in \mathbb{N}$, $h^0(X', -mK_X')=h^0(\mathbb{P}^2,-mK_{\mathbb{P}^2})$, so that $-K_{X'}$ is in fact big, that is $-K_{X'}\geq H$, for some ample $H$.

Hence $\mu^*(K_X)=K_{X'}+E\leq E-H$ is not pseudoeffective, so that it cannot be effective, and the same holds for $K_X$.

Does it make sense?