Existence of a graph G? The Q i am asked is as follows.
Q:Let G=(V,E) be a graph with nine vertices  such that each vertex has a degree 5 or 6. Show that G has at least 5 vertices of degree 6, or at least 6 vertices of degree 5.
first lets bust the middle 5 of6 = 9-5 thus 4of 5 and  6 of 5 = 9-6 hence 3 of 6. 
so they are claiming that 4 vertex of degree 6 and 5 of degree 5 cannot be a graph. and anything else is fine.
clearly $4*6 + 5*5 = 24+25 = 49$  thus $E= 24.5$ which i would like to see you show me half an edge? 
Now i could draw pictures of all the other graphs ( me being lazy i want to find a better way)  or i could ask is the existence of an even sum here sufficient to prove that there exists a simple connected graph? i know that having half an edge is sufficient to disprove the existence  but for a set number of vertices if there exists an even number of edges does a graph need to exist?
EDIT: ( to downvote and comment on duplicate)

My question is for a set number of vertices if there exists an even number of edges does a graph need to exist?

 A: For simple graphs, i.e., those without loops or multiple edges, the complete answer to your final question is given by the Erdős-Gallai theorem: it gives a straightforward necessary and sufficient condition for a finite sequence of positive integers to be the degree sequence of a simple graph. Specifically, if the numbers are $d_1\ge d_2\ge\ldots\ge d_n$, a simple graph with vertices of these degrees exists if and only if
$$\sum_{i=1}^kd_i\le k(k-1)+\sum_{i=k+1}^n\min\{d_i,i\}$$
for $k=1,\dots,n$.
A: I almost overlooked the additoinal question near the end of yourp post:
Given integers $r_1,\ldots ,r_n$ with $r_i\ge 0$ and $\sum r_i=2e$ even, does there exist a graph with $n$ vertices of degrees $r_1,\ldots, r_n$?
Unless you allow multiple edges and loops, the answer is no: 
Firstly, there are certainly a few more necessariy conditions, for example $r_i\le n-1$ as no vertex can be connected to more than all other vertices, and $r_i<e$ as all edges incident with a vertex are distinct. Many more conditions arise, for example $r_1=1$ implies that at most one $r_i$ can equal $n-1$.
A: Let $k_5$ denote the number of degree $5$ vertices and $k_6$ the number of degree $6$ vertices. Since every vertex is of degree $5$ or $6$ we must have
$$k_5 + k_6 = |V| = 9$$
From the handshaking lemma, 
$$5k_5 + 6k_6 = 45 + k_6 = 2|E|$$
This shows that $k_6$ is odd and $k_5$ is even. Therefore, we have the following possibilities
$$(k_5,\ k_6) \in \{(2,7),\ (4,5),\ (6,3),\ (8,1)\}$$
Therefore for all possibilities, we have either at least $5$ vertices of degree $6$ or at least $6$ vertices of degree $5$.
