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Is there any theory which can tell us whether a graph can be made given that it is $m$-regular and has $n$ vertices

I tried making a 6-regular graph with 8 vertices by connecting to chords on the right, and it happens that 1 vertex remains. 8 vertices, 6-regular

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For this question, I'll assume that "graph" means "simple graph".

As pointed out in the comments, for an $m$-regular graph to have $n$ vertices, an obvious necessary condition is $m < n$.

Thus, assume $m < n$.

If $m,n$ are both odd, then no such graph exists since the sum of the degrees is $mn$ which is odd, but for any simple graph, the sum of the degrees is equal to twice the number of edges, which therefore must be even.

If $m,n$ are not both odd, such a graph always exists. Here's one way to construct such a graph . . .

Place the $n$ vertices in the plane, equally spaced around a circle.

Case $(1)\,$: $m$ is even. Connect each vertex to the $m/2$ closest vertices on the left, and also to the $m/2$ closest vertices on the right.

Case $(2)\,$: $m$ is odd. Since $m,n$ are not both odd, $n$ is even. Connect each vertex to the $(m-1)/2$ closest vertices on the left, and also to the $(m-1)/2$ closest vertices on the right. Also connect each vertex to its diametrically opposite vertex.

To clarify, no need to do both left and right connections -- the left connections are accomplished automatically by the right ones (and vice-versa). In my two-way description, I was just describing which vertices are connected to a given vertex. For an actual construction, there's no need to draw an edge that's already been drawn.

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    That's interesting thank you @quasi but won't that be a multigraph? – 2017-01-25
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    No, just a simple graph. No need to draw the same edge twice. I was just describing the connections. – 2017-01-25
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    For simple graph there is one more obvious necessary condition: $m < n$. – 2017-01-25
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    Right -- if $m \ge n$, it can't be m-regular. Agreed! – 2017-01-25
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    I added a picture in the description. For example, that's supposed to be a 6-regular, graph with 10 vertices. I drew 3 edges to the left of M and, 3 edges to the right of M. Out of these 6, one of them will meet L. Also when we do the same from L, one of them meets M, hence isn't it 2 edges between M and L? – 2017-01-25
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    No. I already explained that there's no need to redraw an edge if it already exists. Alternatively, just go _one_ way -- for example, just go right. The left connections to a vertex $v$ will be taken care of by one of the vertices on the left of $v$. – 2017-01-25
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    To make it easier to visualize the symmetry aspect, just draw chords. – 2017-01-25
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    Oh yes, now I get it. Thanks! If possible can you draw one? – 2017-01-25
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    You should do it. Try $n = 12$, $m = 4$ (using chords). – 2017-01-25
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    Okay I tried that, but there's one vertex remaining – 2017-01-25
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52443/discussion-between-swaroop-joshi-and-quasi). – 2017-01-25
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    For each vertex, draw a chord to the two nearest vertices on the right. By symmetry, at the end of he process, how could any vertex look different than any other one? – 2017-01-25
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    No time for chat -- just try the chord version -- and only go right. Left gets taken care of automatically (anyone's left is someone else's right). – 2017-01-25
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    A is supposed to connect to H. A is not supposed to connect to E. – 2017-01-25
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    You have a lot of connections that weren't part of the specification. Only connect to the _closest_ 3 neighbors on the right. – 2017-01-25
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    I got it, thanks! – 2017-01-25