This may be a silly question  but are there interesting results about the invariant: the minimal size of an open affine cover? For example, can it be expressed in a nice way? Maybe under some additional hypotheses?

Consider for simplicity smooth projective varieties defined over $\mathbb C$. In this case, the minimal size equals $n+1$ where $n$ is the dimension of the variety. Proof. Let $M^n$ be such a variety. Take $n+1$ generic very ample divisors $D_0,...,D_n$. Then such divisors don't have a common intersection. At the same time for any $i$ $M^n\setminus D_i$ is affine. In order to show that you do need $n+1$ open affine subvarieties, notice that the complement to an open affine subvariety is a closed variety of dimension $n1$. Now proceed by induction. For projective irreducible varieties exactly the same reasoning should hold, and it can be generalised further I guess 


In fact this is a difficult question in general, I believe. For example, if $X$ is a closed subvariety of ${\mathbb P}^n$, then asking for the minimal number of affine opens that cover the complement ${\mathbb P}^n\setminus X$ is the same as asking for the minimal number of hypersurfaces whose (settheoretic) intersection equals $X$. I believe this question is open in general. 


I think there are some relevant comments in the work of Roth and Vakil. 


The minimal number of open affine sets needed to cover (the coarse moduli space of) $M_{g,n}$ is a famous open problem. The conjecture, which is due to Looijenga if I recall correctly, is that the minimum is given by $g  \delta_{n,0}$ if $g > 1$. Proving this would in one swoop give a proof of Diaz's theorem and several different vanishing theorems on intersection numbers. 


This is not a complete answer by any means, but here are the two most basic arguments. First of all, you have that every projective scheme that can be embedded in $\mathbb{P}^n$ can be covered by $n+1$ open affines, namely the closed subschemes of the affines $U_i = \lbrace z_i \neq 0 \rbrace \cong \mathbb{A}^n$. For a lower bound, think Cechcohomologically: if $X$ can be covered by $k$ affine opens, then $\check{H}^l(X) = 0$ for every $l > k$. If $X$ is Noetherian separated, then Cech cohomology coincides with sheaf cohomology, which indicated that you need at least $\max\lbrace l \;\; H^l(X) \neq 0 \rbrace$ open affines to cover it. 

