Earlier quoted context omitted.
It's not out of the box when I apply error correcting codes to a massive range of problems ;) If you like cool problems, here is one I have not found a simple solution for, but I once saw it in a place where a simple solution should exist. You have an odd number >=3 of items with the property that any time you remove one, there is a way to split the remaining items into two equal count piles of equal weight. Show all…
I'm not sure I get the problem. For number 5, I can use items of weights 3 1 1 1 1 and they are not equal. If you remove 3, you get 1+1=1+1 and in case you remove only of 1s, you can split 3 = 1+1+1.
And this generalizes - if n = 2k + 1, 2k times x and once (2k - 1)x can be split into twice kx after removing the single (2k - 1)x or twice (2k - 1)x after removing one x. For k = 1, i.e. n = 3, (2k - 1)x = x. So there is an infinite family of counter examples for all odd n > 3.