Moving a Weil divisor on a normal surface away from a finite set of closed points Let $Y$ be a normal surface and let $X$ be a closed subscheme of codimension 2, i.e., $X$ is a finite set of closed points.
Let $D$ be a Weil divisor on $Y$. 
Question. Does there exist a Weil divisor $E$ on $Y$ which is linearly equivalent to $D$ and does not go through $X$? (Edit: I do not assume $E$ to be effective.)
Of course, when $D$ does not go through $X$ the answer is yes.
When I say that $E$ does not go through $X$, I mean that the support of $E$ and $X$ are disjoint.
I'm interested in this question in the most general set-up known. For example, $Y$ is an integral noetherian separated excellent normal 2-dimensional scheme. If you prefer sticking to algebraic surfaces, I would be glad to hear about what's possible in that case too.
The motivation for this question comes from an article by Mumford in which he defines an intersection pairing on a normal surface. In this case $X$ is the singular locus of $Y$.
 A: For an arbitrary normal surface Weil divisors cannot be moved: If $X$ is the singular locus of $Y$ then if any Weil divisor could be moved off $X$ it would not intersect the singular locus hence would be a Cartier divisor (since linear equivalence preserves Cartier divisors). So any normal surface on which all Weil divisors are not Cartier gives a counterexample.
A: Not neccesarily. Take $X$ supported on the exceptional divisor $F$ of the blow-up of $\mathbb{P}^2$ at a point and let $D=F$. Then the only divisors linearly equivalent to $D$ is $D$ itself. If however the linear system $|D|$ is base-point free, then what you want should be true.
A: No. Let $Y$ be the quadric cone $\mathrm{Spec} \ k[x,y,z]/(xz-y^2)$ and let $X$ be the singular point $(x,y,z) = (0,0,0)$. Then $\mathrm{Pic}\ Y \cong \mathbb{Z}/2 \mathbb{Z}$, and any divisor not passing through $X$ is trivial. So any divisor which represents the nontrivial class in $\mathrm{Pic} \ Y$, for example $\{ x=y=0 \}$, will not be equivalent to a divisor avoiding $X$. 
Note that this counter-example does not assume divisors are effective, as J.C. Ottem's does. 
