I seem to remember that a K3 surface with big Picard rank always 
has smooth rational curves. 

This question is equivalent to the following question about integral quadratic lattices. Let us call a vector in a lattice ($-2$)-vector if its square is $-2$. Consider a primitive, non-degenerate, indefinite lattice in $H^2(K3, {\Bbb Z})$, which has rank $\geq d$. Does it follow that it always has integer ($-2$)-vectors? I think I can prove that there  exists an indefinite sublattice of rank $11$ which has no ($-2$)-vectors. Is it true that an indefinite  sublattice of rank $>11$  always has ($-2$)-vectors? Is there any sharp bound on ranks of sublattices not admitting ($-2$)-vectors? I would be very grateful for any answers or references to papers where something similar was considered.