The examples you gave can be extended. If a graph has a transitive group then it is weighted regular. Also if the graph such that one point is connected to all other points and if that point is removed the graph remaining has a transitive group then it is weighted regular.

Also there exists graphs that are not weighted regular Look at the graph on 5 points and look at the following graph 1 is connected to all other points for $n$ equal to 2 through 4 $n$ is connected to $n+1$. Then the sum of the weights on 3 and 4 is equal to the sum of the weights on all points of the graph plus the sum of weight one. The sum of the weights on 2 and 5 is the sum the weights on 3 and 4 plus twice the weight of 1. Combining these two results forces the weight on 2 and 5 to be zero and hence not positive so not all graphs are weighted regular.

Also for any graph if a point $x$ connects only to a point $y$ then if $z$ connects to point $y$ it cannot connect to any other point t if it does since the weights of $z$ and $x$ are the same the value assigned to $y$ must equal the sum of the values assigned to $y + t$ and hence $t$ is forced to be zero. This means that the only tree which can be weighted regular is a star and the only forest one with each component a star and if any graph has a component with a vertex connected to only one point it must be a star.

The join of any two weighted regular graphs is weighted regular.
Let the graphs be $G$ and $H$ with $m$ and $n$ points respectively.
Let there weights assigned to each graph so that each node in each graph has the same
weight. Let the weight of each node in $G$ be $w_1$ and the weight of each node
in $H$ be $w_2$. Let the sum of the weights of $G$ be $w_3$ and the sum of the
weights of $H$ be $w_4$. multiply the weights of $G$ by $a$ and $H$ by $b$. Then
the weight of any node of $G$ in the join is $aw_1 + bw_4$ and of any node of H in
the join is $aw_2 + bw_3$
so for the join to be a weighted regular graph $aw_1 + bw_4 = bw_2 + aw_3$ or $a(w_1-w_3)= b (w_2-w_4)$ or $a/b =(w_1-w_3)/(w_2-w_4)$ we
can find such $a$ and $b$ iff $w_1$ is not equal to $w_3$ and $w_2$ is not equal
to $w_4$.But the sum of the weights on all points is not equal to the sum of the weights
on all points on graph so that condition is satisfied.

Now using the comment by Douglas Zare that the convex hull of the rows of the adjacency matrix contains a multiple of the all-1s vector we can find whether any graph is weighted regular in polynomial time. We can convert it to a linear programming problem and there algorithms such as the ellipsoid method that solve linear programming problems in a polynomial amount of time.

In this paper: "On quantum perfect state transfer in weighted join graph" available at the URL: http://arxiv.org/abs/0909.0431 the concept of a weighted regular graph is applied to perfect state transfer on quantum networks.