A slightly different version of the halting problem I got stuck with a question and would like to have a little guidance for the solution.

I need to prove that the next problem is undecidable:


Input - A program
Problem - Does the number of possible inputs for which the program halts is larger than those for which the program won't halt?


I tried to build a reduction that (in case the input is even) halts for every even number, goes to an infinite loop for every odd and runs the program with the input. Or in case the input is odd does the opposite - but it only works if I am able to prove that the number of real odd numbers is equal to the real even numbers.
I won't give you the solution - I think you're on the right track. However I will give a possible way of making this question precise:

Prove that this problem is undecidable:


Input: A (code for a) Turing Machine $M$
Output: $\texttt{Accept}$ if the number of inputs for which $M$ halts is a finite number $\leq n$
$\texttt{Reject}$ otherwise.


Since there are an infinite number of inputs, if $M$ halts for a finite number of them, then it doesn't halt for an infinite number of them.

Of course, it's possible for a machine to halt on an infinite number of inputs, and not halt on an infinite number of inputs (say, halt on even numbers and not on odd numbers). In this case, a hypothetical machine for this problem would $\mathtt{Reject}$.

You might notice some similarity to the halting problem at this point.