I apologise for the basic question; I am reading Huybrecht's Lecture Notes on K3 surfaces, and on p.257 it is mentioned an example of K3 surface with infinitely many smooth rational curves. Precisely, suppose $X$ has an elliptic fibration with a section $C$ of infinite order. Then the multiples $nC$ (wrt addition law on the elliptic curve) are also sections, thus smooth rational curves. Now, being sections, they must be disjoint. But this is what seems puzzling to me: I thought that two disjoint smooth rational curves on a K3 surface should give independent classes in the NeronSeveri group, and so how come there can be infinitely many while $\rho(X)$ is bounded? I guess I am missing something.

1$\begingroup$ To make Sergey's answer even more concrete, try an example such as the Fermat quartic in $\mathbf P^3$. Here an elliptic fibration is given by projecting away from a line on the surface; any other line that is disjoint from the projection centre will give a section. But you can easily find two such lines that are not disjoint from each other. $\endgroup$– Lazzaro CampeottiApr 4, 2017 at 15:53

$\begingroup$ @potentiallydense that is quite illuminating, thank you. Please, turn your comment into an answer $\endgroup$– HeitorApr 4, 2017 at 16:54
3 Answers
(I am just posting my comment as an answer at the OP's request.)
To make Sergey's answer even more concrete, try an example such as the Fermat quartic in $\mathbf P^3$. Here an elliptic fibration is given by projecting away from a line on the surface; any other line that is disjoint from the projection centre will give a section. But you can easily find two such lines that are not disjoint from each other.
They are distinct, but not disjoint. Moreover, intersection numbers tend to infinity.
Here's another way to get a K3 surface with infinitely many smooth rational curves. Let $S$ be the intersection of a $(1,1)$form and a $(2,2)$form in $\mathbb P^2\times\mathbb P^2$, so $S$ will be a K3 surface provided it is nonsingular. The two projections $S\to\mathbb P^2$ are degree 2, so give involutions $i_1$ and $i_2$ of $S$. The composition $f:=i_1\circ i_2$ is an automorphism of $S$ of infinite order. Now you just need to choose your forms so that you can find (a reasonably generic) smooth rational curve $C$ on $S$, then the images $f^n(C)$ for $n\in\mathbb Z$ give infinitely many such curves.
As noted by others, these curve will intersect, and it's a nice exercise to compute the intersection indices. The following may help. Let $D_1=H\times\mathbb P^2$ and $D_2=\mathbb P^2\times H$ be in $\text{Pic}(S)$. (Generally, this is the full Picard group.) Let $\alpha=2+\sqrt3$, and let $E^+=\alpha D_1D_2$ and $E^=D_1+\alpha D_2$ be in $\text{Pic}(S)\otimes\mathbb R$. Then $$ f^*E^+ \sim \alpha^2 E^+ \quad\text{and}\quad f^*E^\sim\alpha^{2}E^. $$