Yeah I got that counter example too, the solutions do it like this: (It makes sense so I don't know where the fallacy is)

Let

denote the position of the

th available item. We must show that

for some i and j. Consider the numbers:

...[1]

and

...[2]

The 90 numbers from [1] and [2] have possible values from 1-89. So by the form of the pigeonhole theorem I stated in the OP, two numbers must coincide. We cannot have two of [1] or two of [2] identical, thus some number in [1] is equal to some number in [2]. Therefore

for some i and j.

Now I understand their working but obviously there exists a counterexample, so where is the fallacy in their working?

Many thanks again!