Graph coloring/partitioning problem I'm interested in the complexity of the following problem: 

Problem $P$: Given an undirected planar graph $G=(V,E)$ and a weight function $w: E \rightarrow \mathbb{Z}$ (so weights can be negative, too), color the vertices in such a way that the sum of the weights of the monochromatic edges (i.e. those between same color vertices) is minimized. Let's call this sum to be the measure for $(G,w)$.

EDIT 1: A major change in the problem definition: $G$ is planar. Previously, I forgot to specify that. And, edge weights are now integers. 

EDIT 2: Below text is part of the original post but it's not very useful as Mikhail Rudoy points out in his answer. 

Here is what I've thought so far: 

Let $k$ be the optimal number of colors, which we do not know beforehand. Suppose an oracle gives us $k$. Then, the problem becomes similar to the minimum edge deletion k-partition (MEDKP) problem, except for the fact that in MEDKP only positive weights are allowed. So, if we modify the weight function by adding a big enough constant $C$ (such that all weights become positive), i.e.  

$w'(e) = w(e) + C,\;\;\;e \in E$

then using $w'$, we exactly have an instance of MEDKP. 

To solve $P$, we solve MEDKP for each feasible value of $k$, i.e. $k=1 \dots |V|$. Let $M_k$ denote the set of monochromatic edges in the solution of MEDKP for $k$. Then, the solution of $P$ would be $ M_{k^*}$  where  $k^* = \underset{k}{\arg\min} \sum_{e \in M_k} w(e)$.

Now, I conclude that the problem $P$  is at least as hard as MEDKP. And since MEDKP is not in APX [1], $P$ cannot be in APX either. 

Is my reasoning correct? Also, independent of my reasoning above, any pointers, hints or comments on the complexity of $P$ is welcome.
Your problem is NP-complete.
Take an instance $G'=(V',E')$ of the (NP-complete) max-cut problem. Let $m=|E'|$.


Create two new vertices $x_1$ and $x_2$, create the edge between $x_1$ and $x_2$, and all edges between $V'$ and $\{x_1,x_2\}$.
The edge $[x_1,x_2]$ receives positive weight $1000m$
All edges $[x_i,v]$ with $i=1,2$ and $v\in V'$ receive negative weight $-2m$
All edges in $E'$ receive positive weight $1$


It can be easily seen that your optimal coloring for this graph


assigns two distinct colors to $x_1$ and $x_2$ (say color 1 to $x_1$ and color 2 to $x_2$)
and colors every vertex $v\in V'$ with one of these colors 1 or 2


Hence, the two-coloring on $V'$ induces a max-cut in $G'$.