I think I might have a solution for this.
We build a graph $G=(V,E)$ consisting of nodes
$V=\{a_1, ..., a_n, b_1, ... , b_n\}$
and edges are consisted two separate sets $E = E_1 \cup E_2$, which are defined as:
$E_1 = \{(a_i,b_i)\}$ and assign $w_i$ as weights to these edges.
$E_2 = \{(b_i,a_j) | b_i < a_j \}$ and assign $0$ as weight to these edges.
Lemma: This graph is a DAG. Because every node is only connected to a node that has a strictly bigger value. So there is an absolute ordering for the graph.
Lemma 2:
Any non-overlapping subset of intervals corresponds to a path in the graph $G$ and vice versa. Moreover the length of the path equals to the sum of the weights of the intervals.
a) subset to path:
Each subset of intervals will be equivalent to choosing their corresponding edges in $E_1$ and then these edges can be connected to each other using the edges in $E_2$. Because edges in $E_2$ have zero weight the total weight of the path is equal to sum of the edges in $E_1$ which is the sum of weights of the intervals.
b) path to subset: Because the graph is a DAG the path doesn't contain two of the same edge. Moreover if two edges $e_1$ and $e_2$ are on the path the cannot overlap. Therefore any path will be equivalent to a non-overlaying set of intervals. Also the sum of the edge weights in $E_1$ is clearly equal to the sum of interval weights in the subset
Therefore, the problem of maximum non-overlaying subset is transformed to maximal path.
In a DAG there is an algorithm of $\mathcal{O}(N+E)$ that computes the longest path. Now since we might have added a quadratic number of edges this will be $\mathcal{O}(N^2)$ solution. The rest of the complexity (including the sorting part) will be lower.
