Nice work, everyone! I like @grayson’s assumption that non-blue-eyed monks are red-eyed. 
But I also agree that any monk without blue eyes is irrelevant to the whole puzzle. They never communicate anything, not even by dying, since we know that monks only die via impeccable logical deduction (c’est la vie). If you don’t have blue eyes you’ll never deduce that you do. So the non-blue-eyed monks might as well be furniture for all the impact they have.
So. There are exactly n blue-eyed monks on the island and no others. The visitor makes their announcement (“there exists at least 1 blue-eyed monk”) on day 1.
(And I made it confusing by referring to dawn but let’s say that if you deduce your eye color you commit suicide that same day. Dawn marks the new day – the point when you deduce that any living monk did not figure out their eye color the previous day. I might edit to original for posterity [DONE] so thanks to everyone above for helping clarify that!)
Here’s my solution!
Hardcore Math Nerd Version of Solution
Theorem: All n monks simultaneously commit suicide on day n.
Proof by induction on n.
Base case (n=1): The one blue-eyed monk learns from the visitor that blue eyes exist and, seeing no other blue-eyed monks, deduces instantly by process of elimination that they must have blue eyes. Thus they commit suicide on day 1.
Induction hypothesis: For an arbitrary blue-eyed monk population of k, they all deduce their eye color and kill themselves on day k. Now consider the case of k+1 blue-eyed monks. They each observe k monks so by the induction hypothesis they all know that (if the blue-eyed monk population is really k) those k monks will all die on day k. But then they don’t. Therefore there must not be only k monks, there must be an additional one, and by process of elimination it must be themself. All k+1 monks reason that way and so all deduce their eye color on day k+1. QED
Version of Solution that Actually Tries to Get You to Believe it
We can build up the solution by starting with the trivial case. Imagine there’s just 1 blue-eyed monk. That monk learns from the visitor that blue eyes exist and, seeing no other blue-eyed monks, deduces instantly by process of elimination that they must have blue eyes. Thus they commit suicide on day 1. Duh.
What about the case of 2 blue-eyed monks? Each monk is looking at the other one going “oh that poor schmuck”. If you see exactly one blue-eyed monk and the visitor says there exist blue eyes then you expect that monk to commit suicide immediately. And then they don’t. Crap. The only reason they wouldn’t commit suicide is that they see another blue-eyed monk. Which, by process of elimination, must be you. Doh.
Now consider n=3 and imagine yourself as one of the blue-eyed monks, hoping the other 2 you see are the only 2. Well, if they’re the only 2 then the previous paragraphs proves exactly what will happen. They’ll die on day 2. But then they don’t. Crap. There must be a 3rd and it must be you. Doh again.
For n=4 it’s the same story. Each monk sees 3 others and we just proved that if there are 3 then they’ll die on day 3. If they don’t then there must not’ve been 3. But they each see 3 so they all simultaneously conclude that the real number is 4 with the 4th being themselves. Womp womp.
What about something like n=100? Well, you can keep walking that reasoning forward as high as you like, so it’s technically true for any n whatsoever!
Discussion
Of course in real reality this is ludicrous. I mean, even aside from the part about blue eyes being evil and calling the conclusion to commit suicide “rational” and all that. You can construct a realistic version where everyone gets a red or blue card on their forehead and you win a million dollar prize for deducing your card’s color, etc. Maybe you need a disincentive to guess. Whatever, we can stick with the monk version and just pretend.
It’s obvious 1 monk will deduce their eye color instantly, it’s pretty solid for n=2, just barely plausible for n=3, and then for n=4 or higher it’s suddenly completely ludicrous. No matter what kind of supergenius everyone is there is just no way that you can genuinely have confidence on day 4 that you have blue eyes just because the 3 monks you see failed to figure out to all commit suicide the previous day. Knowing they’re supergeniuses isn’t enough. You have to know they know they’re all supergeniuses and know that they all know they know that and … I think one more level but I can’t keep that straight because I’m not a supergenius. Eliezer Yudkowsky calls this the Law of Ultrafinite Recursion.
Anyway, I was going to move on to the Fiendish Followup in a separate post, but @narthur and @zedmango have (impressively) spoiled it a bit here. Still, I think I’ll continue undeterred because there’s a lot of fun discussion to have about the Fiendish Followup.
Nonfiendish Followups
But first, let me take a crack at @zedmango’s followups:
- Lying visitor, no blue eyes: Every monk immediately commits suicide. Harsh.
- Deaf monk: Everything happens exactly the same except the deaf monk fails to commit suicide. Everyone else does.
- Partially-known deaf monk: This one stretches credulity even more since I think we want to assume that each of the N monks is incorrectly certain that they’re the only one who knows of the deaf monk. But I believe it doesn’t matter. All the monks except the deaf one still commit suicide on schedule. Knowing about the deaf monk doesn’t help you as long as there are non-deaf clueless monks who’ll be committing suicide on schedule. Knowing they’ll do so tells you your eye color when they don’t.
- N-common-knowledge of deafness: Again, knowing of at least one non-deaf clueless monk is enough to infer your eye color when they don’t commit suicide on schedule. So you (and they) all will commit suicide on schedule. But if the deafness is common knowledge among all the monks then they survive. Because 2 monks (with one deaf) both survive. So a 3rd monk has no expectation of the deaths of the other 2. Etc.
- Statement is "\ge N blue-eyed monks": I think you can just subtract N-1 from the original answer.
- Same statement but false: If the visitor says, for example, there are \ge 2 when there’s only one then the blue-eyed monk can see the visitor is lying and… warn everyone else? Otherwise they’ll all immediately commit suicide, wrongly thinking they’re the 2nd blue-eyed monk.
I guess the lying visitor cases feel less interesting to me. The fascination of the puzzle is what you can infer based on others’ inferences, etc. You can make anyone infer anything (even 1=2) with false premises.
- Odd number: They all die immediately, since they see an even number!
- n/a
- At most N: Nothing happens. They all already know they might have blue eyes.
- n/a
- Exactly N: Instant death.
- n/a
- N isolated groups: The visitor blares the statement over a loudspeaker across the island and each group hears it and whether or not there are other groups out there hearing it is irrelevant. The dynamics in each group play out as if it were the only group.