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.