The net-degree of a vertex $v$, denoted $\text{netdeg}(v)$, in a digraph $G$ is defined by
$$ \text{netdeg}(v)=| ~ \text{outdeg}(v) - \text{indeg}(v) ~| $$
where $\text{outdeg}(v)$ and $\text{indeg}(v)$ are the out-degree and in-degree of $v$.
Show that the edges of any multigraph may be oriented so that $\text{netdeg}(v) \leq 1$ for all its vertices $v$.
My first thought was to order the vertices and alternate edges in and out beginning at the first vertex then move on to the next vertex and orient the remaining edges depending on the net-degree of the remaining vertices, but this approach does not seem to work.
For any graph $G$, perhaps you could add an auxiliary vertex $x$ and make it adjacent to all vertices of odd degree. Call this new graph $G'$. Since there must be an even number of such vertices (so as to keep the degree sum even), $G'$ has an Eulerian circuit. Assign the orientation to $G'$ realizing the circuit starting from $x$. Removing $x$ yields an orientation for $G$ where for each vertex $v\in V(G)$, $netdeg(v)$ is 0 if $deg(v)$ is even and $netdeg(v)$ is 1 if $deg(v)$ is odd.