It doesn't work like that. You want to show that you can't enumerate all reals, so what do you do? You start enumerating all reals. Then you build a new number, whose Nth bit is different from the Nth bit of the Nth real number. You can do that because you have enumerated them, and we have never talked about an infinite N so we are never flipping the last bit of the last number nor anything similar. We are only flipping the Nth bit of the Nth number. Number 819439243? We flip the 819439243th bit. That's all we do.
What do you do then? Well, you compare that new number you created, with the first number, the second number, and so on. And you see that this number is different to all the other numbers.
What does it mean? That it doesn't matter what your enumeration scheme is, there is at least one number that will always be out of your enumeration scheme, and so your enumeration scheme will fail (so this is an absurd also, as we were supposing that this specific enumeration scheme -that could be any enum scheme- was complete). So you don't have as many integers as reals, so QED.
Secondly, if you take the Integer set and say N = the sum of all numbers > 0 you have counted, then you will never reach N as you go to infinity.
As you see, it doens't matter, as in this proof you never need to do the sum of all numbers.