The product of two integers is always an integer. However, the quotient of two integers is not always an integer. This simple fact leads directly to concepts such as "divisibility", "divisors" and "factors", and ultimately to the "prime numbers". Famously, every positive integer can be uniquely written as a product of prime factors. (Unique if you write the factors in ascending order, anyway.)

This whole structure results from the fact that not every quotient has a solution in the integers. If we consider the set of rational numbers instead, suddenly all these interesting properties disappear. (Unless we continue to treat the integers as a "special" subset of the rationals. If we treat all rationals equally, then every quotient is possible, and there's no special structure to see.)

The sum of two integers is always an integer. However, the difference of two integers is also always an integer. So here there is no analogue of the "prime numbers"; there is no unique way of writing an integer as a sum. For a positive integer $n$, there are $n+1$ different ordered pairs of integers who's sum is $n$. And if we allow negative integers, then for every $n$ there are infinity different sums that produce $n$, of an infinite range of lengths.

The rationals have no interesting prime structure, but if we restrict ourselves to just the integers, then such a structure appears. My question: Is there some subset of the integers that we can restrict ourselves to such that a similar structure appears for addition?

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Maybe you could restrict i. the set of initial "additive primes" and ii. the number of compositions, such that every composed number is unique, e.g. staring with $\{1,3\}$ and $1$ composition, you can only get $\{2,4,6\}$... +1 interesting, because I'am searching for a similar set... – draks ... Aug 9 '12 at 8:50

What is interesting, though, is that in most numerical monoids, factorization is not unique! For example, if we consider the set $$M=\langle 2,3 \rangle := \{0,2,3,4,5,...\}=\{2m + 3n \,\vert\,m,n\in \mathbb{N}\cup\{0\}\}$$
Then 2 and 3 are both irreducible (in that they can't be "broken down" any more under addition), but we have $$2+2+2=3+3,$$ and hence two ways to write 6 as a sum of irreducible elements in $M$.