Suppose that $0<\omega_1 < ... < \omega_N $.
Does there exists some $\{A_n \} \subset \mathbb{C}- \{0\}$ such that $\sum_{n=1}^N A_n e^{i\omega_n x} \to 0$ as $x\to \infty$?
This is certainly true for infinite case, since $\int d \omega \ A(\omega) e^{i\omega x} \to 0$ as $x\to \infty$ for integrable function $A$ by Riemann-Lebesgue lemma. I think it is not true for finite case, but I have no idea how I can prove it.
On
The Riemann-Lebesgue lemma hinges on the fact that the values of $f(\omega)e^{i\omega x}$ for $\omega$'s very close to each other cancel out as $x\to\infty$. Since this is not possible for your sum, you cannot expect this to happen.
Indeed, we may invoke the following sifting property to prove this claim:
$$ \lim_{L\to\infty} \frac{1}{L} \sum_{l=1}^{L} e^{i\omega l} = \begin{cases} 1, & \omega\in 2\pi\mathbb{Z} \\ 0, & \omega\notin 2\pi\mathbb{Z}. \end{cases} $$
Now let $f(x) = \sum_{n=1}^{N} A_n e^{i\omega_n x}$ with $0 < \omega_1 < \cdots < \omega_N$ and $A_1, \cdots, A_N \in \mathbb{C} \setminus\{0\}$. Our aim is to prove that $f(x) \not\to 0$ as $x\to\infty$. To this end, assume otherwise that $f(x) \to 0$ as $x \to \infty$. Then by the sifting property and the Stolz-Cesaro Theorem,
$$ A_N = \lim_{L\to\infty} \frac{1}{L} \sum_{l=1}^{L} f(2\pi l/\omega_N) = \lim_{l\to\infty} f(2\pi l/\omega_N) = 0, $$
a contradiction.
Let $f(x) = \sum_{n=1}^N A_n e^{i\omega_n x}$. Then ${1 \over k}\int_k^{2k} |f(x)|^2 dx= {1 \over k}\int_k^{2k} f(x)\bar{f(x)} dx=$ $${1 \over k}\int_k^{2k}\sum_{m,n = 1}^N A_m \bar{A_n}e^{i(\omega_m - \omega_n)x}dx$$ $$=\sum_{m,n = 1}^N A_m \bar{A_n}{1 \over k}\int_k^{2k} e^{i(\omega_m - \omega_n)x}dx$$ The terms where $m \neq n$ go to zero as $k \rightarrow \infty$, so you have $$\lim_{k \rightarrow \infty} {1 \over k}\int_k^{2k} |f(x)|^2 = \sum_{n=1}^N |A_n|^2$$ This is incompatible with $\lim_{x \rightarrow \infty} f(x) = 0$ since the average of $|f(x)|^2$ over intervals $[k,2k]$ would also have to go to zero as $k \rightarrow \infty$.