First things first - there are an infinite number of both even numbers and odd numbers.
It's important to realize that $\infty$ (infinity) is not a number. Therefore it doesn't really make sense to talk about the "number" of even or odd numbers, or to write statements like $E_{\rm even}+1$, because that's assuming that $E_{\rm even}$ is a number that you can sensibly add $1$ to.
However, perhaps surprisingly it does make sense to ask if there are more even numbers than odd numbers. That is, you can compare two infinite quantities, or compare a finite quantity and an infinite quantity, even if you can't meaningfully add and subtract infinite quantities.
They way we define more, less and the same for infinite quantities is as follows. For two collections $A$ and $B$ (say $A$ are the even numbers and $B$ are the odd numbers) we say that
If you can associate every item in $A$ with a unique item in $B$, and vice versa, then $A$ and $B$ are the same size.
If you can associate every item in $A$ with a unique item in $B$, but not vice versa, then $B$ is bigger than $A$.
If you can associate every item in $B$ with a unique item in $A$, but not vice versa, then $A$ is bigger than $B$.
In your case, you can associate every even number $n$ with the odd number $n+1$, and you can associate every odd number $m$ with the even number $m-1$ (assuming 0 is even) so therefore there are just as many odd numbers as even numbers.
This can lead to seemingly paradoxical results, because e.g. you can associate every whole number $n$ with the even number $2n$, and every even number $m$ with the whole number $m/2$, so there are just as many even numbers as whole numbers, even though the even numbers are a subset of the whole numbers.
Best Answer
Hypothesis: "$n$ is a positive integer"
Conclusion: "the sum $1+2+\cdots+n$ equals $\frac{n(n+1)}{2}$"
and the given statement is: if hypothesis then conclusion.