Part 1) Begin by putting smallest number > 1 as say, 4 in set B and continue filling in the numbers in A and B as follows, at each point place the number $max(a+b)+1$ where $a\in A$ and $b\in B$ into either of the sets A or B such that $4k+r \in A$ for all $1 \leq r \leq 3$, if $4k$ is in $A$ and place only multiples of 4 in $B$. Below are 2 examples patterns of filling in the next higher multiple of 4 as below: A = ${0,1,2,3, 8,9,10,11 \ldots}$ B = ${0,4,16,20,32 \ldots}$ A = ${0,1,2,3, 8,9,10,11, 16,17,18,19 \ldots}$ B = ${0,4,24,28,48 \ldots}$ See part 2 for more detailed explanation of the pattern. Part 2) Since $0$ can only be expressed as $0 + 0$, clearly $0\in A$ and $0\in B$. WLOG let $1\in A$. We shall prove that $B$ contains only multiples (not all multiples) of some integer $k>1$. Let $k$ be the smallest positive integer such that $k\in B$. This would imply that all $i \in A$ such that $1 \leq i \leq k-1$. $--> (1)$ Now to prove that $B$ contains only multiples of $k$ consider the following $2$ cases: Case $1$: Consider a multiple of $k$ say $pk$ in $B$: Here we cannot have $pk+r \in B$ for all $1 \leq r \leq k-1$. This is because this will invalidate the uniqueness of representing a number $n = pk+r$. It can be represented in $2$ different ways as follows. We can choose $pk$ from $B$ and $r$ from $A$ (from $(1)$ above) as well as $pk+r$ from $B$ and $0$ from $A$. Also, for the same reason we cannot have $pk+r \in A$ for all $1 \leq r \leq k-1$ if $pk$ is in $B$. Case $2$: Consider a multiple of $k$ say $qk$ in $A$: Here we cannot have $qk+r \in B$ for all $1 \leq r \leq k-1$. This is because this will invalidate the uniqueness of representing a number $m = qk+k$. It can be represented in $2$ different ways as follows. We can choose $qk+r$ from $B$ and $k-r$ from $A$ (from $(1)$ above) as well as $qk$ from $A$ and $k$ from $B$. Thus we must have $qk+r \in A$ for all $1 \leq r \leq k-1$, if $qk$ is in $A$. Thus, the above $2$ cases establish that $B$ can only have multiples of $k$, while $A$ must have all $k-1$ numbers immediately following any multiple of $k$ that exists in $A$. Note that both $A$ and $B$ are infinite sets and we can choose a multiple $hk$ for $h > 1$ to be part of either sets.

KRIS17 wrote: Part 1) Let $A$ be the set of all even multiples of $4$, such that each multiple of $4$ is followed by next $3$ numbers that immediately follow this multiple. i.e. ${0, 1, 2, 3, 8, 9, 10, 11, 16, 17, 18, 19 \ldots }$ Let $B$ be the set of all odd multiples of $4$, after including $0$. i.e. ${0, 4, 12, 20 \ldots }$ It can be verified that $A$ and $B$ satisfy the conditions of the problem. This unfortunately does not actually work: take $n=13$, which is $1+12$ and $4+9$ in your sets.

benstein wrote: KRIS17 wrote: Part 1) Let $A$ be the set of all even multiples of $4$, such that each multiple of $4$ is followed by next $3$ numbers that immediately follow this multiple. i.e. ${0, 1, 2, 3, 8, 9, 10, 11, 16, 17, 18, 19 \ldots }$ Let $B$ be the set of all odd multiples of $4$, after including $0$. i.e. ${0, 4, 12, 20 \ldots }$ It can be verified that $A$ and $B$ satisfy the conditions of the problem. This unfortunately does not actually work: take $n=13$, which is $1+12$ and $4+9$ in your sets. That was indeed a good catch my friend! I have edited my example for Part 1). Part 2 is still correct!

a) $A$ is the set of all nonnegative integers whose binary representation has the digit $1$ in even places only. $B$ is the set of all nonnegative integers whose binary representation has the digit $1$ in odd places only.