In keeping with my answer to your Computer Science question, here's an answer that only involves 3 sheets of paper and doesn't involve showing me any numbers for the two-possibility case. I can't come up with a good answer for the three-possibility case, but I've given an answer that's not completely bad.
We have to assume you can't write anything on a paper except the allowed numbers (1, 2, and possibly 3), and any paper with a given number is indistinguishable from other papers with the same number.
Two Possibilities (1, 2)
The numbers 1 and 2 can be substituted for any other integers provided one is even and one is odd.
For the sum to be even, the two papers must contain the same number. So the proof is simply that the papers match: either both are 1, or both are 2.
So you write a number on a third sheet of paper that isn't the first number. If the existing sheets are both 1, the third sheet is 2. Otherwise the third sheet is 1.
Now, you prove the third sheet doesn't match the first sheet. You hand me both sheets. I ensure I never lose track of which is which, but I hide them from you so you can't tell while shuffling.
I show you (and only you) both sheets. You now know which is on the left and right. I hide them from you, and either leave them in place, or switch them left to right, randomly. I again show you each sheet, so you can tell left from right. You tell me whether I switched them or left them the same.
If you can consistently tell when I switched them or didn't, I know they're different. Otherwise, the test fails.
Since I kept track of when I switched them, I know which one is the third sheet. I place the first sheet to the side, then take the second sheet. I repeat the test to prove you know which is the second sheet and which is the third.
If you can consistently tell me which is the second and which is the third sheet, I know they're different. Since you were only allowed to write one of two numbers, and neither of the first two sheets matches the third, the first two sheets must match each other.
Three Possibilities (1, 2, 3)
We can substitute 1, 2, and 3 with any other integers provided two are odd and one is even.
Unfortunately, my answer to the Computer Science question won't completely work here.
That answer only shows the two numbers match. Here, it's also acceptable if the numbers are 1 and 3. So we have to prove neither of the papers has the number 2, or both have the same number.
First, we can simply test using the above method.
You give me four papers, $P_1, P_2, P_3, P_4$. The papers $P_2, P_3, P_4$ must have the numbers 1, 2, and 3, but I don't know which is which. $P_1$ + $P_2$ is claimed to be even.
Using the standard method, I prove that each pair of papers in [$P_2, P_3, P_4$] is different. I test you with $P_2\neq P_3$, $P_2\neq P_4$, and $P_3\neq P_4$. This ensures you have one of each number.
Now, I prove that $P_1$ is neither $P_3$ nor $P_4$ using the standard method, which tells me $P_1$ must be $P_2$.
If this proof succeeds, we're done. The result is guaranteed to be even. However, failure here isn't a total failure since there were multiple ways the result could be even.
Sadly, there's no way to find the answer without giving away some extra information.
If the two numbers don't match, any test proving the sum is even automatically tells us the numbers are 1 and 3.
You can write the number 2 on a paper, $P_5$, then use the standard method to prove that neither $P_1$ nor $P_2$ matches $P_5$. This proves that both are 1, both are 3, they're 1 and 3, or they're 3 and 1.
This never tells us the exact value of either paper, but I don't really like it, given the spirit of "zero information" protocols. I'd prefer a method that never distinguishes between
matching papers and papers aren't 2
but I can't think of such a method.
Real World (Lack of) Applicability
I'm unable to come up with any scenario where these "proofs" really work. As a puzzle, we can magic away some of the rules and that's fine. But it doesn't work in a practical setting.
The original test, that two objects are definitely different, is clear-cut. I can't tell two things apart, but you can. That's potentially useful, and is immediately convincing.
In this test, I require proof that you can't cheat. It would be trivial to make a small mark on one paper, or write the number in way you can see the difference but I can't, or just write completely different numbers than agreed upon. Since I'm not allowed to examine the papers, I have no way of showing that it's even likely you can't cheat.
So I'm required to either take your word for it, in which case I can just skip the entire test. Or take the word of a trustworthy third-party, in which case I can just have them verify on my behalf.
It might work as a magic trick of some kind. Use cards from brand-new card decks, and have an audience member verify the cards aren't tampered with. But magicians can easily acquire trick decks that are sealed, so that's not really convincing. And your average audience member isn't going to have to logical aptitude to be any more convinced by this proof than you just claiming you're right.