What is the maximum number of vertices given a number of edges? What is the largest number of vertices in a graph with 35 edges if all vertices are of degree at least 3?
I thought up of a solution but there are some questions I could not find.
Solution 
Maximum degree of a graph >= Sum of degree of individual vertices
2E >= deg(V1) + deg(V2) + ... + deg(Vn)
2 * 35 >= 3 + 3 + ... + 3            ...(I), 
70 >= 3n
23.33 >= n or 23 >= n
So the largest number of vertices are 23, with the given constraints.
Question
If we are considering the degree of each of the vertices to be 3, then shouldn't the number of vertices be even i.e 22 ?(since number of odd degree vertices in a graph are even)
 A: You don't know that all vertices have degree $3$, only that their degree is at least $3$. Therefore your argument that there cannot be an odd number of vertices doesn't work.
Your argument that $70 \ge 3n$ still works in essentially the same way; the initial inequality should just be written as
$$ 2E = \text{sum of degrees} \ge 3+3+\cdots+3 = 3n $$

You still need to prove that $23$ is indeed possible, for example, by showing a concrete graph with $23$ vertices (of degree $\ge 3$) and $35$ edges.
One way to make such a graph would be:


*

*Start with your favorite 3-regular graph on $22$ vertices (for example, the edge graph of an 11-sided prism). It has $33$ edges, so we need to add one vertex and two edges.

*Take one of the edges, and bisect it by a new vertex. That gets us up to $34$ edges. The new vertex so far has degree only $2$, but fortunately we still need to add one edge.

*Add an edge from your new vertex to one of the existing vertices (which it isn't already connected to). This gives the new vertex degree $3$ and the other endpoint degree $4$, which is allowed.
