Suppose $G$ is a finite group. Let $\chi$ be the character of some $\mathbb{C}G$-module with the property that $\chi(g)$ is an even integer for every $g \in G$. Is it true that $\chi/2$ defined by $\chi/2 (g):=\chi (g) /2$ is the character of some $\mathbb{C} G$-module?
Let $V$ be a $\mathbb{C}G$-module and $\mathcal{B}$ be a basis. Then the map $\chi:G \rightarrow \mathbb{C}$ defined by $\chi(g)=trace([g]_{\mathcal{B}}$) is the character of $V$.
I think it is equivalent to ask: Does there exist another $\mathbb{C}G$-module $W$ and a basis $B'$ of $W$ such that $\frac{1}{2}trace([g]_{\mathcal{B}})=trace([g]_{\mathcal{B'}})$ for every $g \in G$ if $trace([g]_{\mathcal{B}})$ is an even integer?
No. Let $G$ have even order $n$ and let $\chi$ be the character of the regular representation. Then $\chi(e)=n$ and $\chi(g)=0$ for $g\ne e$. So $\chi/2$ is integer valued. But $\chi/2$ is not a character. The regular representation includes the trivial representation with multiplicity $1$. A representation with character $\chi/2$ would have to include the trivial representation with multiplicity $1/2$.