Find the number of subsets of {1, 2, . . . , n} with an odd number of elements. Find the number of subsets of {1, 2, . . . , n} with an odd number of elements.

I am looking at an example where they are asking for an even number of sets. Do the number of even and odd sets have to be equal? I am confused :/
If n=0, the answer is 0, since there is only one subset of the empty set (itself), and that has even size.

If n>0, then there are $2^{n-1}$ ways to choose a subset $S$ of $\{1,2,...,n-1\}$. For each of them, exactly one of $S, S \cup\{n\}$ has odd size. 

Thus, there are $2^{n-1}$ such sets.
Say $S_n=\{1,2,\dots,n\}$. First assume $n$ is odd. Then $E\subset S_n$ has an odd number of elements if and only if $S_n\setminus E$ has an even number of elements. So the number of odd subsets is the same as the number of even subsets, hence $2^n/2=2^{n-1}$.

Now suppose $n$ is even and $E\subset S_n$. Let $$F=E\cap S_{n-1}.$$ For $E$ to be odd there are two possibilities: $F$ is even and $n\in E$ or $F$ is odd and $n\notin E$. So the number of odd subsets of $S_n$ is the same as the number of subsets of $S_{n-1}$, again $2^{n-1}$.