Solution: et $G = (V,E)$ be an undirected graph with $n$ edges. Let $V_e$ and $V_o$ denotes the set of vertices of even degree and the set of vertices of odd degree of $G$, respectively. We recall the
Handshaking Theorem.
Let $G = (V,E)$ be an undirected graph with $n$ edges. Then
\[
\sum_{v \in V} \deg(v) = 2n,
\]
where $\deg(v)$ denotes the degree of the vertex $v$.
Using the Handshaking theorem,
\begin{align*}
& 2n = \sum_{v \in V} \deg(v) = \sum_{v \in V_e} \deg(v) + \sum_{v \in V_o} \deg{v} \\
\implies & 2n - \sum_{v \in V_e} \deg(v) = \sum_{v \in V_o} \deg{v}.
\end{align*}
Since, $\sum_{v \in V_e} \deg(v) $ is even, the sum $\sum_{v \in V_o} \deg{v}$ must be even. But each of the terms $\deg(v)$ is odd, so there must be even number of elements in $V_o$. Hence, there are eben number of vertices with odd degree.