Prove or disprove (Eulerian Graphs) Which part of the following question is correct?
A graph is eulerian if and only if the maximum number of edge-disjoint paths between any two vertices of this graph is an even number. ( a graph is eulerian if it has a circuit which contains all of its edges)
I personally think that if a graph is eulerian, then the maximum number of edge-disjoint paths between any two vertices of this graph is an even number. But I think the other side of this problem is not necessarily correct .
My idea for proving the first part is that if we consider that the maximum number of edge-disjoint paths between two vertices like $u$ , $v$ is odd. For example name these paths as $p_1,\dots,p_{2k+1}$ . now omit the edges in these cycles : $up_1vp_2u$ , $up_3vp_4u$ , $\dots$ , $up_{2k-1}vp_{2k}u$. The graph still remains eulerian. But there exist just one path between $u$ and $v$ and this is a contradiction.  
 A: Suppose that $G$ is an Eulerian graph, $s$ and $t$ are any distinct vertices of $G$. By edge connectivity version of Menger’s theorem, the maximum number of edge-disjoint $s$-$t$-paths is equal to the minimum number of edges whose removal disconnects $s$ from $t$. Let $G’$ be the graph $G$ with the minimal number of removed edges providing $s$ and $t$ belong to different connected components of $G’$, $S$ and $T$, respectively. Then the set of the removed edges consists of edges between $S$ and $T$ in $G$, otherwise we can put back a removed edge, keeping the vertices $s$ and $t$ disconnected, contradicting the minimality of the set of removed edges. But number of the edges between $S$ and $T$ in $G$ equals the number of changes of a component ($S$ or $T$) of a vertex, going along an Eulerian circuit in $G$, which is an even number.  
The other side implication can fail, for instance, for a graph without edges or for a union of two vertex-disjoint cycles. But maybe the implication holds for connected graphs $G$. I tried to construct a counterexample, but it seems to be not so simple. Anyway, it is easy to show that it should be 2-edge-connected and have at least four vertices of odd degree.
