A graph is *1-planar* if it can be drawn on the plane such that each edge is 
crossed at most once. 

Let $G$ be a 1-planar bipartite graph with $n~(n > 4)$ vertices and $m$ edges. Karpov \[1\] showed that $m \ge 3n − 8$ holds for even $n \ge 8$ and $m \ge 3n − 9$ holds for odd $n \ge 7$.

\[1\]. D. V. Karpov. Upper bound on the number of edges of an almost planar bipartite
graph. J. Math. Sci., 196:737–746, 2014.

So the minimum degree of any  1-planar bipartite graph is at most $5$. Here is my question.

 - Construct a 5-regular bipartite 1-planar graph.

I note that $5n\le2(3n-8)$ impling that $n\ge 16$.  Maybe we will find such graph with $16$ vertices.


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The problem comes from planar bipartite graphs. Any planar bipartite graph has minimum degree at most 3. The smallest order of 3-regular planar bipartite graph is 8; see the below graph.
[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/Ccp1l.png