A nationwide math contest in Germany recently came up with a task that I found beautiful to explain, because of two points.
You can get an idea, what the proof is, without applying mathematically accurate theory and this intuitive proof is most likely the right way.
At any given point of this intuitive proof, you can chime in and ask yourself: But how would I say this in mathematical terms? When you find these terms, eventually you get the proof you were looking for.
So here you go: Lea gets the task to write down 2014 numbers. These numbers have to fulfill a specification. For every set of three numbers from that whole set, the arithmetic average of these three must also be within the whole set of $2014$ numbers.
Your task is to proof, that Lea has to write down the same number $2014$ times. Every set of 2014 numbers with any variation in it would not fulfill the specifications.
So since we are talking about layman maths here, I'll go with the intuitive way. We have to find a reason, why choosing a set with different numbers would violate the specifications and we have to proof that always taking the same number would not violate them. The later one is rather easy. Take any arbitrary number three times. The arithmetic average will be the same number, which is in your set already. That wasn't too bad, right?
But what about sets with not all the same numbers? We are mathematicians, so we'll just do what we always do: Chop the problem into pieces we can solve. The first piece is where we have two equal numbers and one other number in our set. Let's assume, the single number is bigger than the two equal numbers. What will that do to our arithmetic average? Right, it will be below the middle between the bigger and the smaller number. We can write that arithmetic average down and specifications are ok. But now we have created another set of three numbers. The single, big number (I'll call it a), one of the two equal numbers (that would be b) and the arithmetic average of a, and b (I'll call that one c). So now we would have to also add the arithmetic average of a, b and c. A quick sketch will show you, that this new number is also slightly below the middle between a and b.
And like that we will always have to add a new number. The arithmetic average of a, b and the new number will never reach the middle. Something, that you can also verify with a few sketches. So we would have to add infinitely more numbers, but we wanted only 2014. Apparently, no two numbers can be equal.
So what if all numbers are different? There is one special case. Let's call our numbers a, b and c again. If b is equally far away from a and c (so b could be 3, a could be 1, then c would be 5). In that case, b is the arithmetic average. But we have to have 2014 different numbers. As soon as we add a fourth number d, it's spoiled. d could be 7, to be still in a distance of 2 to c, but then the set a, b and d would not contain its own arithmetic average. So we know, that within a set of 2014 numbers, we would have sets of three, where these three numbers don't include their own arithmetic average, no matter what we do.
And now we look back at our idea about the set with two equal numbers. We see: As soon as we have a bigger and a smaller number and the number in between those is not exactly in the middle, we can once again start with our endless construction of arithmetic averages. We always replace the number between the bigger and the smaller one by the arithmetic average of the three and we can never reach the middle, but it will always get closer to the middle (thus be another number).
And as I said, making this proof mathematical will not alter it. It will be all the same, but with more equations and sequences. Since we excluded the option of making anything infinite, it is correct as it stands here. This one made me realize: Proofs are not the miracles or the magic they seemed to be for me during high school. Of course, there are hard proofs (and things you can't proof, there is a proof for that), but often you only have to think clearly and to chop the problem into the right pieces.