Maximizing sum edge weights I am wondering if the following problem has a name, or any results related to it.

Let $G = (V,w)$ be a weighted graph where $w(u,v)$ denotes the weight of the edge between $u$ and $v$, and for all $u,v \in V$, $w(u,v) \in [-1,1]$. The problem is to find a subset of vertices that maximizes the sum of the weights of the edges adjacent to them:
$$\max_{S \subseteq V} \sum_{(u,v) : u \in S\ \textrm{or}\ v\in S} w(u,v)$$
 Note that I am counting edges both that are inside the subset and that are outside the subset, which is what distinguishes this problem from max-cut. However, even if both $u$ and $v$ are in $S$, I only want to count the edge $(u,v)$ once (rather than twice), which is what distinguishes the objective from merely being the sum of the degrees.

Note that the problem is trivial if all edge weights are non-negative -- simply take the whole graph!
FWIW, your problem is hard to approximate within a multiplicative factor of $n^{1-\epsilon}$ for any $\epsilon&gt;0$. 

We show that below by giving an approximation-preserving reduction from Independent Set, for which the hardness of approximation is known.

Reduction from Independent Set

Let undirected graph $G=(V,E)$ be an instance of Independent Set.
Let $d_v$ denote the degree of vertex $v$ in $G$.
Let $n$ be the number of vertices in $G$.

Construct edge-weighted graph $G'=(V',E')$ from $G$ as follows.
Give each edge in $E$ weight 1.
For each non-isolated vertex $v\in V$, 
add $d_v-1$ new edges, each with weight $-1$,
to $d_v-1$ new vertices.
For each isolated vertex $v\in V$,
add one new edge of weight 1 to a new vertex.

(Note: each new vertex (in $G'$ but not $G$) has exactly one neighbor,
which is in $G$.)

Lemma. $G$ has an independent set of size $k$ iff
$G'$ (as an instance of your problem) has a solution of value at least $k$.

Proof.  Let $S$ be any independent set in $G$.
Then, since the vertices in $S$ are independent in $G'$,
the value of $S$ in $G'$ (by your objective)
is 
$$\sum_{v\in S} d_v - (d_v-1) ~=~ |S|.$$

Conversely, let $S$ be a solution for $G'$ of value at least $k$.
Without loss of generality, assume $S$ contains no new vertices.
(Each new vertex $v'$ is on a single edge $(v',v)$.
If $v$ was not isolated in $G$, then the weight of the edge
is $-1$, so removing $v'$ from $S$ increases the value of $S$.
If $v$ was isolated, then the weight of the edge is 1,
so removing $v'$ from $S$ and adding $v$ maintains the value of $S$.)

Without loss of generality, assume that $S$ is an independent set in $G$.
(Otherwise, let $(u,v)$ be an edge such that $u$ and $v$ are in $S$.
The total weight of $v$'s incident
edges in $G'$ is $d_v - (d_v-1) = 1$, so the total weight of $v$'s incident
edges other than $(u,v)$ is at most zero.
Thus, removing $v$ from $S$ would not increase the value of $S$.)

Now, by the same calculation as at the start of the proof,
the value of $S$ is $|S|$.  It follows that $|S| \ge k$. QED

As an aside, you might ask instead for an additive approximation,
of, say, $O(n)$ or $\epsilon m$.  

It seems possible to me that for your problem even deciding whether 
there is a positive-value solution could be NP-hard.