An algebraic vector bundle is trivialized by open sets. How many does one need? Consider an algebraic vector bundle $E$ on a scheme $X$. By definition there is an open cover of $X$ consisting of open subsets on which $E$ is trivial and if $X$ is quasi-compact, a finite cover suffices. The question then is simply: what is the minimum number of open subsets for a cover which  trivializes $E$ ? Now this is silly because the answer obviously depends on $E$ ! If $E$ is trivial to begin with, the cover consisting of just $X$ will do, of course, but if you take $\mathcal O(1)$ on $\mathbb P^n_k$ you won't get away with less than $n+1$ trivializing open subsets . Here is why.
Suppose you have $n$ open subsets $U_i\subset \mathbb P^n_k$ over which $\mathcal O(1)$ is trivial. Take regular nonzero sections $s_i\in \Gamma(U_i,\mathcal O(1) )$ and extend them rationally to $\mathbb P^n_k$. Each  such extended rational section $\tilde {s_i}$ will have a divisor $D_i$ and the complements  $\tilde U_i= X\setminus |D_i|$,  $(U_i\subset \tilde U_i)$, of the supports of those divisors  will give you a cover of $\mathbb P^n_k$ by $n$ affine open subsets trivializing $\mathcal O(1)$. But this is impossible , because $n$ hypersurfaces in $\mathbb P^n_k$ cannot have empty intersection.
This, conversations with colleagues and some vague considerations/analogies  have led me to guess ( I am certainly not calling my rather uninformed musings a conjecture)  that the following question might have a positive answer:

Is it true that on  a (complete) algebraic variety of dimension $n$ every vector bundle is trivialized by some cover consisting of at most $n+1$ open sets?          

For example, the answer is indeed yes for a line bundle  on a (not necessarily complete) smooth curve $X$: every line bundle $L$ on $X$ can be trivialized by two open subsets .
Edit Needless to say I'm overjoyed at Angelo's concise and brilliant positive answer. In the other direction ( trivialization with too few opens to be shown impossible) I would like to generalize my observation  about projective space. So my second question is:

Consider a (very) ample line bundle $L$ on a complete variety $X$ and a rational section 
  $s \in \Gamma _{rat} (X, L) $. Is it true that its divisor $D= div (s)$ has a support $|D|$ whose complement  $X\setminus |D|$ is affine ? Let me emphasize that the divisor $D$ is not assumed to be effective, and that is where I see a difficulty.

 A: This is true if we assume that the vector bundles has constant rank (it is clearly false if we allow vector bundles to have different ranks at different points). Let $U_1$ be an open dense subset of $X$ over which $E$ is trivial, and let $H_1$ be a hypersurface containing the complement of $U_1$. Then $E$ is trivial over $X \smallsetminus H_1$. Now, it is easy to see that there exists an open subset $U_2$ of $X$, containing the generic points of all the components of $H_1$, over which $E$ is trivial (this follows from the fact that a projective module of constant rank over a semi-local ring is free). Let $H_2$ be a hypersurface in $X$ containing the complement of $U_2$, but not containing any component of $H_1$. Then we let $U_3$ be an open subset of $X$ containing the generic points of the components of $H_1 \cap H_2$, and let $H_3$ be a hypersurface containing the complement of $U_3$, but not the generic points of the components of $H_1 \cap H_2$. After we get to $H_{n+1}$, the intersection $H_1 \cap \dots \cap H_{n+1}$ will be empty, and the complements of the $H_i$ will give the desired cover.
[Edit]: now that I think about it, you don't even need the hypersurfaces, just define the $H_i$ to be complement of the $U_i$.
A: 
This is an answer to Georges' updated question at the end of his post.

An equivalent formulation of the question is the following:

Question 
  Let $L$ be an ample Cartier divisor on a projective scheme $X$ and suppose there exist effective divisors $D_1, D_2$ such that $L\sim D_1-D_2$. Then is it true that   $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)$ is affine? 

I think this is true in some cases, but not in general.

Claim 1 The answer to the question is YES if $X$ is a projective curve.

Proof 
Both $D_1$ and $D_2$ are effective and hence ample and similarly so is $A=D_1+D_2$. Clearly 
 $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)=X\setminus {\rm supp}\, A$, which is affine. $\square$

Claim 2 There are many examples for smooth projective varieties for which there exists $L, D_1, D_2$ as above such that  $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)$ is not affine. In fact, this happens on any smooth projective surface containing a $(-1)$-curve. 

Remark I am pretty sure one does not need smoothness and there are also singular examples. (Actually the example below only needs one smooth point.)
Proof
Let $Y$ be an arbitrary projective variety (reduced) of dimension at least $2$ and $H$ an effective (very) ample Cartier divisor on $Y$. Let $\sigma : X\to Y$ be the blow up of a smooth point $p\in Y$ that is not contained in $H$ and let the exceptional divisor of $\sigma$ be $E\subset X$. 
Then for some $m>0$ positive integer, $L=m\sigma^*H-E$ is ample. (I suspect that most people know this, but if you need a hint for this statement, an explicit estimate on $m$ can be found in Lemma 2 of this answer to another MO question.)
Now let $D_1=m\sigma^*H$ and $D_2=E$. Notice that by the choice of the point that was blown up, $D_1$ and $D_2$ are disjoint. It follows that  $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)\simeq (Y\setminus {\rm supp}\, H)\setminus \{p\}$. Furthermore, since $H$ is ample on $Y$, it follows that $Y\setminus {\rm supp}\, H$ is affine, and hence 
$(Y\setminus {\rm supp}\, H)\setminus \{p\}$ is not. $\square$
It is actually true, that for any line bundle there always exists a rational section for which the complement of its divisor is affine. 

Claim 3 
  Let $L$ be an arbitrary Cartier divisor on a projective scheme $X$. Then there exist effective very ample divisors $D_1, D_2$ such that $L\sim D_1-D_2$.

Proof
Choose an arbitrary ample Cartier divisor $A$ on $X$. For large enough $r_1\gg 0$ 
$L+r_1A$ is basepoint-free by the definition (or one of the basic properties depending on what you choose as definition) of ampleness. Then for an even larger $r\gg r_1$ we may assume that $L+rA$ is both basepoint-free and ample and hence very ample and also that $rA$ is very ample as well. Now choose $D_1=L+rA$ and $D_2=rA$. $\square$
And we get as an easy consequence:

Corollary
  With the notation of Claim 3, we may choose $D_1$ and $D_2$ such that
    $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)$ is affine. 

Proof
Replace $D_1$ and $D_2$ with general members of their complete linear systems. Then we may assume that they do not have a common component and hence ${\rm supp}\,(D_1+D_2)={\rm supp}\,D_1 \cup {\rm supp}\,D_2$. Since $D_1+D_2$ is also ample, this proves that claim. $\square$
