Question: If $X$ is a projective surface and $U$ is an open affine subset of $X$, then is it true that $X \setminus U$ is the support of an (effective) ample divisor on $X$?

Background: I was reading Goodman's paper "Affine open subsets of algebraic varieties and ample divisors" which considers this same question for general varieties. Here is what I understand so far:

If $\dim X = 1$, then the answer to the question is always affirmative. Indeed, if $X$ is a complete curve and $S$ is any finite set of points on $X$, then there is an effective ample Cartier divisor on $X$ which has support $S$ (this is Proposition 5 of the paper, and a straightforward application of the Nakai-Moishezon criterion of ampleness).

For $\dim X = 2$, Theorem 2 of the paper states that the answer is positive if each point of $X\setminus U$ is factorial (i.e. its local ring is a UFD). Actually he proves it only assuming that $X$ is complete (i.e. a priori not necessarily projective) and as a corollary he proves Zariski's theorem that if all the singularities of a complete surface $X$ are contained in an affine open subset, then $X$ is projective.

He presents two examples (of Hironaka and Zariski) where $X$ is a non-singular projective $3$-folds, but $X\setminus U$ is not the support of any ample divisor.

In general he proves (in Theorem 1) that if $X$ is complete then a Zariski open subset $U$ of $X$ is affine iff the complement of (the isomorphic image of $U$) in a blow-up $X'$ of $X$ along a closed subscheme $F$ not meeting $U$ is the support of an effective ample Cartier divisor on $X'$.

For $\dim X \geq 3$, he gives a criterion (in Theorem 3) for when the answer to the question is positive.

As far as I can see, he does not mention anything about the status of the question (i.e. whether if there is a counter-example or not) for general projective surfaces. Therefore I ask it here. I would expect the answer to be negative, but can not think of any examples. For me particularly interesting would be the case when $X$ is normal.

** Edit: ** As the example of Jason Starr in the comment shows: The answer is * negative * even for normal surfaces (see my comments for an attempt of proof). I wonder what happens if $X$ is rational. In any event, I would gladly accept Jason's answer if he writes one. (And I would also greatly appreciate any answer/remark about the rational case.)