Recently I was thinking and came up with a conjecture that goes as follows:
Conjecture:
There exists a $\Omega$ such that $$\Omega = \Bigg\{A_i \ \Bigg| \ \forall i,j:i\not=j, \ A_i\cap A_j=\emptyset , \ \bigcup_{i} A_i=\Bbb N:\sum_{k\in A_i} k=13^i \Bigg\}$$ where $0\not \in \Bbb N$
You can do it inductively in the "naive" way: Set $A_1=\{1,12\}$, $A_2=\{2,13^2-2\}$,$\ldots$, $A_{11}=\{11,13^{11}-11\}$, $A_{12}=\{13,13^{13}-13\}$, $A_{13}=\{14,13^{14}-13\}$,...
You have, however, to show that this sequence satisfies the properties you want. I think one elementary way to do it is the following: We want to define $A_i$ inductively with the properties
Define $A_1=\{1,12\}$. Given $A_1,\ldots,A_n$ satifying the three properties above, let $k=\min(\mathbb{N}\setminus(A_1\cup\cdots\cup A_n))$, and define $A_{n+1}=\{k,13^{n+1}-k\}$. Properties 1.,2. and 4. are immediate. As for property 3., we need to show that $13^{n+1}-k$ does not belong to some $A_i$, $i\leq n$. Since each $A_i$ has two elements, $A_1\cup\cdots\cup A_n$ has at most the $2n$ first elements of $\mathbb{N}$, so $k\leq 2n$. But then $13^{n+1}-k\geq 13^{n+1}-2n>13^n\geq 13^i$ for all $i\leq n$, so property 2. implies $13^{n+1}-k\not\in A_i$< as desired.