Need hint for bipartiteness proof
Let's do the case $N = 1$.
Suppose that $G$ is a connected graph, and assume first that $G$ is bipartite, with bipartition $X,Y$. Let $X_1,X_2,Y_1,Y_2$ be the two copies of the two parts in $G'$. Note first that the edges in $G'$ either connect $X_1$ to $Y_2$ or $X_2$ to $Y_1$, and so $G'$ has at least two connected components.
Conversely, any path from $X$ to $Y$ in $G$ gives rise to a similar path in $G'$ from $X_1$ to $Y_2$, implying (after a short argument) that $X_1 \cup Y_2$ form a connected component; and similarly for $X_2 \cup Y_1$. Thus $G'$ has exactly two connected components.
Now suppose that $G$ is a connected non-bipartite graph. As before, the vertices in $X_1 \cup Y_2$ are all in the same connected component of $G'$, as are the vertices in $X_2 \cup Y_1$. We will show that $X_1$ is connected by a path to $X_2$, implying that $G'$ is connected.
Since $G$ is not bipartite, there is some odd cycle in $G$. Let $x$ be a vertex on this odd cycle. Lifting this cycle to $G'$, we obtain a path from $x_1$ to $x_2$.