Here is an expansion of joro's answer.
Claim. $K_{n, n+1}$ is obvious if and only if $n+1$ is even.
Proof. If $n+1$ is even, we can add a perfect matching on the vertices on the right to obtain a $(n+1)$-regular graph. Conversely, if $n+1$ is odd, I claim that $K_{n,n+1}$ is not obvious. Obviously (bad pun), we cannot delete a matching to create a regular graph, since deleting any edge decreases the degree of both a vertex on the left and a vertex on the right. Thus, we must add a matching $M$, and no edge of $M$ can be incident to a vertex on the left, since they already have degree equal to $n+1$. It follows that $M$ must be a perfect matching of the vertices on the right, but this is impossible since $n+1$ is odd.
The same proof actually gives a slightly bigger class of graphs that are not obvious. First, we can extend to complete multipartite graphs. That is, $K_{n, \dots, n, n+1}$ is not obvious when $n+1$ is odd. Next, $K_{n, \dots, n, n+1}+M$ is not obvious when $n+1$ is odd, and $M$ is a matching with all endpoints in the $n+1$ part of the partition.