Two visualization problems that many people have trouble with (no props allowed).
Imagine 3 vertical parallel lines intersecting 3 horizontal parallel lines. How many squares are formed?

Imagine a coin rotating completely around the edge of a coin of the same denomination. How many rotations does it make?

Here are two problems that I presented on Fluther that I am happy to say were solved, but which cause most people difficulties.

How many numbers can be chosen from 1 to 100 such that no number is divisible by any other. Some people have trouble with this even with the following hint: Any number can be expressed as O*2^n, where O is odd.

Imagine people arranged in rows and columns. Have the people in each column arrange themselves in order of height, front to back. Now have each row of people arrange themselves in order of height left to right. Are the people in each column still arranged in order of height. Why or why not?

This one is based on a computer programming algorithm.
You have four coins all of different weights. Using a balance, what is the minimum number of weighings required to guarantee being able to find both the heaviest and the lightest? How can the method be generalized for any even number of coins?

Some people find this really easy and some (me included) have trouble with it
Take a deck of cards. Turn 20 of them face up and shuffle into the deck. Have someone count off 20 cards from the top and hand you the remainder of the deck. What can you do to make sure that the number of face up cards that you have equals the number of face up cards in the 20 cards.

@LostInParadise GA. Also, in the one about the numbers 1–100 that are not divisible by each other, can you explain what you mean by “divisible”? Does it mean that the two numbers have no factors in common, or that one does not go into the other an integer amount of times? (in short, are 15 and 20 considered divisible?) The reason I am asking is that my first thought was simply numbers 51–100, which would be valid by one definition but invalid with the other.

51 to 100 is correct. The problem is to prove that the most you can get is 50 such numbers. A more advanced variation is to find the smallest solution.

I messed up on the card problem. It should be what should be done so that the number of face down cards is the same as in the 20 cards.