If we assume that the participants are numbered, each participant asks himself "is there some X and Y such that #X does not know that #Y knows that #101 sees blue?".

And just to be clear, we can apply this to 3 blues.

Imagine 3 blues and 5 browns and 1 green.

BL1(#X) sees 2 blues, and looks at one of them (#Y) and knows that he sees at least 1 blue, and because #Y sees at least one blue, #X can reason that #Y also knows that guru sees at least one blue.

So if this is truly the basis of the reasoning, it has to work at 3 blues.

Then go through all the numbers and for each number imagine the participants asking themselves "is there some X and Y such that #X does not know that #Y knows that #1 sees blue/brown/green?" — Michael

X knows that everyone knows that guru sees blue at 3 blue. But we've already established that 3 can't leave on the third day.

You're trying to address the problem, but this is a deduction puzzle, and your deduction has a false premise. The premise that's false is 99 blue eyed people would leave on the 99th day.

But for me to show you that's false, I would have to show you that it's false that 98 people would leave on the 98th day.

And to prove that's false, I would have to prove to you that it's false that 97 people would leave on the 97th day.

And so on.

That's a lot.

But here's the deal- you keep counting down, 99 98 97... eventually you get to 3. And we know 3 don't leave on the third day.

It's easier to talk about small numbers than big numbers.

I'm very interested in that number. — flannel jesus

Then go through all the numbers and for each number imagine the participants asking themselves "is there some X and Y such that #X does not know that #Y knows that #1 sees blue/brown/green?"

I'm just addressing the problem in the OP, and I think that what I say in that post above shows that the blues and browns can and will leave on the 100th day having deduced their eye colour even without the Guru having said anything.

I explain it in the first part of the post above:

If we assume that the participants are numbered, each participant asks himself "is there some X and Y such that #X does not know that #Y knows that #101 sees blue?".

Given that there are 201 participants, there are 40,401 possible combinations, so it's unfeasible for us to list them all, although our perfect logicians will be able to.

But we can make a start.

I'm #1 and I see 99 blue (#2-100), 1 green (#101), and 100 brown (#102-#201).

I ask myself:
Does #2 know that #101 sees blue? Yes; #2 knows that #101 can see blue #3.
Does #2 know that #1 knows that #101 sees blue? Yes; #2 knows that #1 and #101 both see blue #3.
Does #2 know that #2 knows that #101 sees blue? Yes; #2 knowing what #2 knows is a tautology.
Does #2 know that #3 knows that #101 sees blue? Yes; #2 knows that #3 and #101 both see blue #4.
Does #2 know that #4 knows that #101 sees blue? Yes; #2 knows that #4 and #101 both see blue #3.
Does #2 know that #5 knows that #101 sees blue? Yes; #2 knows that #5 and #101 both see blue #3.
Does #2 know that #6 knows that #101 sees blue? Yes; #2 knows that #6 and #101 both see blue #3.
...

I'm fairly certain the answer is always going to be "Yes", and so that there is no X and Y such that #X does not know that #Y knows that #101 sees blue.

So, returning back to the previous steps, I deduce that (1) is true

"is there some X and Y such that #X does not know that #Y knows that #1 sees blue?"

I don't know what the minimum number of participants must be for the answer to this question to be "no", but by induction it appears that the answer to the question is "no" when applied to the problem in the OP, which is the only thing I'm addressing.

In order to get started, so that the failure of anybody to leave is meaningful, all this must be known. And, for Michael solution to work, all this must be known too. Only a truth telling guru communicating to everyone that indeed there is a blue can cut through this recursive epistemic conundrum. — hypericin

I'm not convinced that it need be recursive. It's sufficient that each person knows that each person knows that green sees blue.

If we assume that the participants are numbered, each participant asks himself "is there some X and Y such that #X does not know that #Y knows that #101 sees blue?".

Given that there are 201 participants, there are 40,401 possible combinations, so it's unfeasible for us to list them all, although our perfect logicians will be able to.

But we can make a start.

I'm #1 and I see 99 blue (#2-100), 1 green (#101), and 100 brown (#102-#201).

I ask myself:
Does #2 know that #101 sees blue? Yes; #2 knows that #101 can see blue #3.
Does #2 know that #1 knows that #101 sees blue? Yes; #2 knows that #1 and #101 both see blue #3.
Does #2 know that #2 knows that #101 sees blue? Yes; #2 knowing what #2 knows is a tautology.
Does #2 know that #3 knows that #101 sees blue? Yes; #2 knows that #3 and #101 both see blue #4.
Does #2 know that #4 knows that #101 sees blue? Yes; #2 knows that #4 and #101 both see blue #3.
Does #2 know that #5 knows that #101 sees blue? Yes; #2 knows that #5 and #101 both see blue #3.
Does #2 know that #6 knows that #101 sees blue? Yes; #2 knows that #6 and #101 both see blue #3.
...

I'm fairly certain the answer is always going to be "Yes", and so that there is no X and Y such that #X does not know that #Y knows that #101 sees blue.

So, returning back to the previous steps, I deduce that (1) is true, and can make use of the subsequent counterfactuals to determine whether or not I am blue, which I will deduce on either the 99th day (if I'm not) or the 100th day (if I am):

1. As of right now everyone has come to know, through some means or another, that everyone knows that #101 sees blue
2. If (1) is true and if I do not see blue then I am blue and will leave this evening
3. If (1) is true and if I see 1 blue then if he does not leave this evening then I am blue and will leave tomorrow evening
4. If (1) is true and if I see 2 blue then ...
...

And it bears repeating (if any reader missed the previous comment), that even though as a practical matter (1) is true in counterfactual scenarios (2) and (3) only if someone says "I see blue" isn't that someone must say "I see blue" in every counterfactual and actual scenario for (1) to be true and for this reasoning to be usable.