Is it possible to have $\sum_{g\in G} \chi(g)=\frac{1}{2}$? 
Let $V$ be a finite-dimensional vector space over a field $F$ and let $\rho : G \to GL(V)$ be a representation of a group $G$ on $V$. The character of $\rho$ is the function $\chi : G \to F$ given by
$${\displaystyle \chi(g)=\operatorname {Tr} (\rho (g))}$$
where $\operatorname {Tr}$ is the trace.

Is it possible there exists a finite group $G$ and a character $\chi$ of $G$ such that
$$\sum_{g\in G} \chi(g)=\frac{1}{2}?$$
I believe not, but I have not been able to come up with any demonstration of this fact. And is the fact that $G$ is finite important? I mean, supposing $G$ to be an infinite group, would the answer to this question be the same?
Note: Assuming the field is $\mathbb{C}$.
I believe this is not true (finite case), because the sum of elements in row of character table is a positive integer(*), but in this case, the characters in the table are irreducible, can I conclude the same for a regular character? Since a regular character can be represented as a sum of irreducible characters.
(*) But in this case, I don't have that result up to the point of this question.
 A: Yes it is possible.
Let $G$ be cyclic of order $2$, and let $F$ be the finite field of order $3$. Let $\rho$ be the trivial representation.
Then $\sum_{g \in G}\chi(g) = 2 =_F 1/2$.
Incidentally, specialists use Brauer characters rather than ordinary characters when dealing with representations over fields with finite characterits (modular representation theory). One of the reasons for doing this is to avoid this kind of undesirable behaviour.
A: Note that
$$ \sum_{g\in G}\chi(g)=|G|\cdot \langle \chi,1_G\rangle,$$
where $1_G$ is the trivial character. Thus it is always a non-negative integer. And if you aren't allowed to assume that, then you have to essentially prove that, by showing that your sum counts the number of fixed points in the representation.
A: Perhaps this is helpful. In addition to the remarks above: if $\chi$ is a character of $G$, then $\chi(g)$ is an algebraic integer for all $g \in G$ and since sums of algebraic integers are again algebraic integers, the sum cannot be the rational $\frac{1}{2}$. Now let's look at a proof without an appeal to algebraic integers. 
If $\chi$ would be a class function rather than a character, the situation is different. For example if $\chi$ would be the irreducible character of degree $2$ of $S_3$, and $\lambda$ the non-trivial linear character of $S_3$, then $|\chi|$ is a class function but not a character. In fact one can easily calculate from the character table that
$$|\chi|=\frac{2}{3}1_G + \frac{2}{3}\lambda + \frac{1}{3}\chi.$$
Note that $[|\chi|,|\chi|]=[\chi,\chi]=1$, so $|\chi|$ has norm $1$ as with the irreducible characters, still it is not a character! In fact, every class function $\varphi$ can be expressed as $\varphi=\sum_{\chi \in Irr(G)}a_{\chi}\chi$, with $a_{\chi} \in \mathbb{C}$. And here $\varphi$ is a character if and only if $a_{\chi} \in \mathbb{Z}_{\geq 0}$. Now in your question, you are looking at $|G| \cdot [\chi,1_G]=|G|a_{1_G}$, which hence is always an integer. So fractions will never appear.
So this leaves us with the following question: if $\varphi$ is a class function (not necessarily a character) is always $|G|[\varphi,1_G] \in \mathbb{Z}_{\geq 0}$? Well, this is not true in general. A counterexample can be found with $G=PSL(2,7) \cong GL(3,2)$, the simple group of order $168$. Let $\chi$ be one of its irreducible characters of degree $3$, and put $\varphi=|\chi|$. Then $\varphi$ is a class function but a calculation shows that $|G|[\varphi,1_G]=66+48\sqrt{2}$.
