Simultaneous Countably Infinitely Many Hats

First, we show that they can guarantee at most finitely many will be wrong. The mathematicians apply an equivalence relation to the set of all sequences of red and blue hats. They agree that two sequences are equivalent if they differ in at most finitely many places (I will not prove here that this is an equivalence relation, but it is very easy to check for yourself). By the axiom of choice, they may agree on a representative sequence from each equivalence class. When it is time to guess, the mathematicians look ahead of them and are able to tell in which equivalence class the actual sequence of hats belong. They then guess as if the actual sequence was the agreed upon representative - for example, if I am mathematician 100, and I know that the 100th hat in the representative sequence corresponding to the equivalence class containing the actual sequence is red, then I will guess red. Of course, by definition of the equivalence relation at most finitely mathematicians will be incorrect.

Now we must show that they cannot put an upper bound on the number of mathematicians who will be incorrect. Suppose otherwise, i.e. they can guarantee no more than $n$ mathematicians will be wrong (take $n$ to be sharp). Add a new mathematician (say, mathematician zero) to the back of the line behind mathematician one. This does not change the problem, so $n$ remains a sharp upper bound on the number of mathematicians who could guess correctly. However, the original mathematicians might as well not even know mathematician zero exists, as s/he can convey no new information - therefore, $n$ mathematicians not including mathematician zero might still guess incorrectly. Therefore, we must be guaranteed that mathematician zero will guess correctly. But of course, mathematician zero's knowledge does not change at all if their hat color changes, so the probability that they will guess correctly is 1/2. By contradiction, no upper bound can be given.