Here is an idea of how one (let's say I) might try to construct such an example. I think this idea, if it works, might qualify under rule #2. Also, it is possible that the answer Matt gave on MSE help fill the gap in this. Even though this is not a complete answer I think it still might be useful.

So, let's say that $X$ is a smooth projective variety and $\mathscr L$ an ample line bundle on $X$ such that $H^1(X,\mathscr L^{-1})\neq 0$. This can obviously happen only in positive characteristic by Kodaira vanishing, but it *can* happen there. I think Raynaud was the first to give examples of this (in 1978?) and Ekedahl showed that this can even happen with $\mathscr L=\omega_X$ in characteristic $2$. Others gave various examples for this to happen, I think maybe including Mukai.

Anyway, suppose that such an $X$ comes in a polarized family (that is, there is a relatively ample line bundle on the total space that restricts to $\mathscr L$ on $X$) where the general fiber *does not* have satisfy this (for any power of the line bundle that's the deformation of $\mathscr L$). I am not sure if this is an outrageous expectation, but I am kind of thinking that the failure of Kodaira vanishing is *special* and so a *general* deformation will not be a counter-example to Kodaira vanishing, in other words Kodaira vanishing holds on it. One potential problem I see with this is that it is possible that such an $\mathscr L$ would perhaps not deform, so I could not reasonably take a "general" deformation. Well, actually if one uses $\mathscr L=\omega_X$ as in Torsten's example, then it definitely deforms! I am sure that Torsten will read this and tell me why I am wrong. :)

In any case, what I need is just that $\dim H^1(X,\mathscr L^{-1})$ would jump and this still seems an easier task than the original since $\mathscr L$ is undetermined, so one has a little more freedom.

So, if we have that, then we're in business. Take a general section of a high power of the global line bundle that restricts to $\mathscr L$ and take the finite cover it determines. This finite cover has the property that the cohomology of $\mathscr O$ of the fibers of this cover over the original base is the sum of the cohomologies of the negative powers (up to the degree minus one) of the deformations of $\mathscr L$. The assumption implies that either the original family gave you an example or there will be a jump in this one.

isas a precise mathematical statement (beyond elimination of quantifiers), which likely reflects my own ignorance. See point 3 at the end of <a href="math216.wordpress.com/2011/07/21/seventeenth-post/… post</a> for more of my thoughts on this. As I state there, perhaps I should ask this on MathOverflow. – Ravi Vakil Jul 25 '11 at 16:50