A property of graphs containing a vertex of odd degree Let $G$ be a connected graph with at least one vertex of odd degree. Is it possible to assign $\pm 1$'s to each edge such that the for any vertex, the absolute value of the sum of the  numbers on edges going through that vertex is at most 1 ?
 A: If $G$ has $2n$ vertices of odd degree ($n\gt0$), then $E(G)$ can be decomposed into $n$ edge-disjoint trails (*); each trail has two of the odd-degree vertices as its endpoints. Label the edges of each trail with alternating signs.
Now suppose $G$ is a connected graph with no vertices of odd degree, i.e., an Eulerian graph. In this case the condition "the absolute value of the sum of the numbers on edges [incident with each] vertex is at most $1$" means that exactly half the edges at each vertex are positive. If $G$ is an Eulerian graph with an odd number of edges, the condition cannot be satisfied; if $G$ is an Eulerian graph with an even number of edges, it can be satisfied by choosing an Euler circuit and labeling the edges with alternating signs.
The above discussion is for finite graphs. More generally, if $G$ is a locally finite graph (meaning that each vertex has finite degree), the following are equivalent:  
(1) it is possible to assign a positive or negative sign to each edge, so that at each vertex the number of positive edges and the number of negative edges differ by at most one;  
(2) every component which is finite and Eulerian has an even number of edges.
(*) The following theorem and proof are copied from p. 30 of Introduction to Graph Theory, Second Edition, by Douglas B. West.

1.2.33. Theorem. For a connected nontrivial graph with exactly $2k$ odd vertices, the minimum number of trails that decompose it is $\max\{k,1\}$.Proof: A trail contributes even degree to every vertex, except that a non-closed trail contributes odd degree to its endpoints. Therefore, a partition of the edges into trails must have some non-closed trail ending at each odd vertex. Since each trail has only two ends, we must use at least $k$ trails to satisfy $2k$ odd vertices. We also need at least one trail since $G$ has an edge, and Theorem 1.2.26 implies that one trail suffices when $k=0$.It remains to prove that $k$ trails suffice when $k\gt0$. Given such a graph $G$, we pair up the odd vertices in $G$ (in any way) and form $G'$ by adding for each pair an edge joining its two vertices, as illustrated above. The resulting graph $G'$ is connected and even, so by Theorem 1.2.26 it has an Eulerian circuit $C$. As we traverse $C$ in $G'$, we start a new trail in $G$ every time we traverse an edge of $G'-E(G)$. This yields $k$ trails decomposing $G$.

