Let $X$ be a projective variety and let $L$ be a line bundle on $X$. Suppose for all locally free sheaves $M$ on $X$,
$
H^i(X,{L^\*}^{\otimes r} \otimes M)=0
$ for $i<\dim X$ and $r$ sufficiently big. 


Does it follow that $L$ is an ample line bundle? Here $L^\*$ denotes the dual of $L$.


This is of course clear if $X$ is smooth using Serre duality, but how is it in general?


After reading Laurent Moret-Bailly and Karl Schwede's comments, below I changed the condition '$M$ coherent' to '$M$ locally free'.



Ample line bundle and Duality
As was pointed out in the comment above by Laurent Moret-Bailly, you can't choose any $M$. Let me try to answer a slightly different question which I hope is close to what you intended.


**Edit:** This should also answer the revised question, in the negative.


The answer is no, ample line bundles do not satisfy the condition you want, even for $M = \mathcal{O}\_X$ (or more generally for $M$ a vector bundle). In other words, suppose what you wanted was true, then if $L$ satisfied the condition it would be ample (this is what you wanted), but as I'll show below, this will also imply that $L$ cannot be ample.


If $X$ is Cohen-Macaulay, then what you want holds (see the version of Serre-duality in Hartshorne). 


Suppose now that $X$ is not Cohen-Macaulay (CM) but its non-CM-locus is isolated at a point $z \in X$ (for example, if that point looks locally analytically like the cone over an Abelian variety). Take $M = \mathcal{O}\_X$ and suppose that 


By assumption, $H^i(X, L^{-r} ) = 0$ for all $i < d = \dim X$ and all $r \gg 0$.
However, 
$$H^i(X, L^{-r}) = \mathbb{H}^{d - i}(X, L^r \otimes \omega\_X^{\bullet}[-d])^{\vee}.$$
Here $\omega\_X^{\bullet}[-d]$ is the dualizing complex of $X$ shifted over by $d = \dim X$. Being CM is equivalent to $\omega\_X^{\bullet}$ just being a sheaf (and not a complex). 


Now we wish to compute $\mathbb{H}^{d - i}(X, L^r \otimes \omega\_X^{\bullet}[-d])^{\vee}$. This is done using a Grothendieck-spectral sequence (see for example Weibel's homological algebra). Since $r \gg 0$ and $L$ is ample, this implies that the spectral sequence degenerates at the $E^2$ stage (in other words, degenerates immediately). The upshot of this is that
$$
\mathbb{H}^{d - i}(X, L^r \otimes \omega\_X^{\bullet}[-d]) = H^0(X, L^r \otimes h^{d-i}(\omega\_X^{\bullet}[-d]) ).
$$
Where the $h^{d-i}(\omega\_X^{\bullet}[-d])$ just means that $(d-i)$th cohomology of the shifted dualizing complex. Since $X$ was not Cohen-Macaulay with isolated non-CM-locus, some these are supported at the non-CM point $z \in X$. In other words, some of the $L^r \otimes h^{d-i}(\omega\_X^{\bullet}[-d])$ for $i < d$, are skyscraper sheaves (finite dimensional vector spaces even), so some of the $\mathbb{H}^{d - i}(X, L^r \otimes \omega\_X^{\bullet}[-d]) \neq 0$.


