For $n$ real numbers $x_1,x_2,...,x_n$ we can define their modulus sum as: $$ f(x_1,x_2,...,x_n)=\left(\sum_{k=1}^{n}x_k\right) \mod 1,$$ where $z=(a \mod 1)$ is a number in the interval $[0,1)$ such that $(a-z)$ is integer. I want to find an extension to sum an infinite sequence of numbers. More precisely, does there exists a function $f:\mathbb{R}^{\infty}\to[0,1)$ (the input of $f$ is a sequence of real numbers) such that if we fix all the numbers in the sequence $(x_1,x_2,...,x_{n-1},x_{n+1},x_{n+2},...)$ except for $x_n$ then there is an identity: $$f(x_1,x_2,...,x_n,...)=\left(c+x_n\right) \mod 1$$ for some constant $c$ (that may depend on the rest of the sequence, but does not depend on $x_n$). This should hold for every $n\in\mathbb{N}$.
I was trying to prove the existence of such function using Zorn's lemma, but for now without success.
The answer to your question is yes, and in fact one can construct a function satisfying a stronger condition than the one you give, removing any mention of modulo $1$. On ${\mathbb R}^{\infty}$, let $\sim$ be the equivalence relation defined as follows : $(x_n) \sim (y_n)$ iff $x_n-y_n$ is zero except for finitely many values of $n$. Let $C=\frac{{\mathbb R}^{\infty}}{\sim}$. By the axiom of choice, we can choose for each $c\in C$ a sequence $z^c \in C$.
Let $x=(x_n)$ be any sequence in ${\mathbb R}^{\infty}$, let $c$ be the $\sim$-equivalence class of $x$, and let $(z_n)=z^c$. Put
$$ f((x_n)_{n\geq 1})=\sum_{n\geq 1}x_n-z_n \tag{1} $$
It is easy to see that (1) defines a function that's compatible with $\sim$, in the sense that
$$ f((x_n)_{n\geq 1})-f((y_n)_{n\geq 1})=\sum_{n\geq 1}x_n-y_n, \ \textrm{whenever} \ (x_n) \sim (y_n). $$