How to show that some problem is in NP and that it's NP-complete
When performing a reduction it is important to keep in mind exactly what you are doing. In particular you are trying to reduce from a known hard problem (here subset sum, $\mathsf{SS}$) to your problem (here great subset sum, $\mathsf{GSS}$) such that yes-instances of the hard problem map to yes-instances of your problem and no-instances map to no-instances.

As mentioned in the comments, if you simply divide each number in $S$ by two (and also divide $n$ by two) you do not get an instance of $\mathsf{GSS}$; it might not be the case that $\frac{n}{2} \geq \frac{N}{2}$.

Please think about the problem a bit more before looking at the solution below.


   Let our $\mathsf{SS}$ instance be $S = \{s_1, ..., s_k\}$ with $\sum_{s_i \in S} s_i = N$ and target $n$. We construct the following $\mathsf{GSS}$ instance: $S' = \{s_1, ..., s_k, N\}$ with target $n' = N + n$. Observe that this is a $\mathsf{GSS}$ instance since $n' \geq \frac{1}{2}\sum_{s \in S'} s = N$. It remains to show that yes-instances map to yes-instances and no-instances map to no-instances. If some subset $A \subseteq S$ adds up to $n$ then $A \cup \{N\}$ adds up to $N + n$. If not then no subset of $S'$ adds up to $n'$ since $n' &gt; N$ so this subset must contain $\{N\}$ and a subset of $S$ which adds up to $n$.