I am looking for some (simple and well-known) sufficient conditions on a quasiprojective variety to be Kobayashi hyperbolic.
I realize that in this generality it may be a complicated (maybe even hopelessly naive) question, so less generality is perfectly OK.
A bit more specifically, I have read in some paper a comment like "this is the complement in a (complex) projective space of a certain number of hyperplanes, so it is Kobayashi hyperbolic". Why is that? What is the statement that seems to be so classical that the author did not include it?
I'm certainly far from knowing the subject well. However, what follows quite easily is the following:
Suppose $X$ is Kobayashi hyperbolic. Then $X$ cannot contain any rational curves, and any rational map $f:A\to X$ from an abelian variety (or more generally any compex torus) must be constant.
The reason is simply that the Kobayashi pseudodistance is zero on the complex plane. The converse statement is an open problem linked to the Lang conjecture, which suggests that a variety $X$ is of general type if and only if $X\neq U$, where $U$ is the closure of the union of the images of all rational maps from an abelian variety to $X$.
Further, it is known that if $X$ is compact then it is Kobayashi hyperbolic if and only if there are no non-constant maps $\mathbb{C}\to X$. (Brody's theorem)
The criterion you read about is probably the following: Suppose $X=\mathbb{P}^{n} \backslash D$ where $D$ is the union of $2n+1$ hyperplanes in general position. Then $X$ is Kobayashi hyperbolic.
The question whether the complement of a single generic hypersurface of some large enough degree is hyperbolic, is also an open problem I think.
All this (and much more) can be found in the book by Kobayashi.