if $|S|<2$ we must have $|S|=1$ ($S=\{a\}$) we have for $n=3$: $\forall\{a_{i}\}_{i\in\mathbb{N}}\in S=\{a\}:\sum_{k = 0}^{3}a_{k}x^{k}=a(x^{3}+x^{2}+x+1)$ isn't irreducible in $\mathbb{Z}[X]$. if $|S|\ge 2$ we take $a,b\in S$ such that $a=a'd,b=b'd,a'>b',gcd(a',b')=1$ we take $\{a_{i}\}_{i\in\mathbb{N}}\in S:\ a_{0}=a,\forall i>0,a_{i}=b$ then we have $\forall x\in\mathbb{Z}:\forall n\in\mathbb{N}\frac{\sum_{k=0}^{n}a_{k}x^{k}}{d}\equiv a'\ (mod\ b')\not\equiv 0\ (mod\ b')$ then $\forall n\in\mathbb{N},\forall x\in\mathbb{Z}:\sum_{k=0}^{n}a_{k}x^{k}\neq 0$ so $\forall n\in\mathbb{N}:\ \sum_{k = 0}^{n}a_{k}x^{k}$ is irreducible in $\mathbb{Z}[X]$.

Obviously above solution for $\mid S\mid \geq 2$ is wrong It’s the right time for “BUMP”.

yes @above, you're rigth \bump

Iranian National Olympiad (3rd Round) 2007 wrote: Prove that for a set $S\subset\mathbb N$, there exists a sequence $\{a_{i}\}_{i = 0}^{\infty}$ in $S$ such that for each $n$, $$\sum_{i = 0}^{n}a_{i}x^{i}$$ is irreducible in $\mathbb Z[x]$ if and only if $|S|\geq2$. Here's a proof for the $|S| \ge 2$(the 'if' direction') Choose primes $$\begin{align*} p_1=a_0 10^2+a_1 \\ p_2=a_0 10^3+a_1 10^2+a_2 \\ . \\.\\.\\.\\p_n=\sum_{i = 0}^{n}a_{i}10^{i} \end{align*}$$ which by Dirchlet's theorem exist, hence by Cohn's irreduciblity Criterion $$\sum_{i = 0}^{n}a_{i}x^{i}$$ is irreducible.$\blacksquare$