Suppose on the contrary $g(x) := \sum_{i=1}^n f_i(x)-\sum_{i=1}^n f_i(x-a_i) \ge 1\,,\, \forall x\in \mathbb{R}$. Let $N$ be a positive integer. Denote $ J= \{ (j_1,j_2,\dots,j_n) : j_k \in \mathbb{N}\,,\, 1\leq j_k \leq N\,,\, k=1,2,\dots,n \}\,,\, a=(a_1,a_2,\dots,a_n)$. For any $j\in J$ by $j\cdot a$ we mean $j_1a_1+\dots+j_na_n$. Now, consider: \[A_N := \sum_{j\in J} g(j\cdot a) \,\,\,\,\,\,\,\, (*)\] Obviously, we have $A_N \geq N^n$. On the other hand, all $f_i(j\cdot a)$, for which $ 1 \leq j_k < N$ are canceled, and there remain only terms $f_i(j\cdot a)\,, j=(j_1,\dots,j_n)$, for which some $j_k$ is either $0$ or $N$. Let's denote: \[J_0= \{(j_1,j_2,\dots,j_n) : 0 \leq j_k \leq N\,, k=1,2,\dots,n\,,\, \exists k, 1\leq k \leq n\,,\, j_k=0 \}.\] \[J_N= \{(j_1,j_2,\dots,j_n) : 0 \leq j_k \leq N\,, k=1,2,\dots,n\,,\, \exists k, 1\leq k \leq n\,,\, j_k=N \}.\] The above note yields: \[A_N \leq \sum_{i=1}^n \sum_{j\in J_0\cup J_N} |f_i(j\cdot a)| \leq 2 n (N+1)^{n-1} M \] where, $M$ is such that all $|f_i|$ are bounded by $M$ Thus, we obtain: \[N^n < 2M n (N+1)^{n-1} \] But letting $N\to \infty$, we get a contradiction.

Take the set $X = \{ s=\displaystyle\sum_{i=1}^n c_ia_i : s\in [-N,0], c_i \in \mathbb{Z}, c_ia_i<0 \}$ and sum over all $x \in X$. For big $N$, we get a contradiction.