Cute problem! Define the following triangle so that the $i^{\text{th}}$ row has $\lceil\tfrac{i}{2}\rceil$ entries, and the $j^{\text{th}}$ column consists of the multiples of $j$ starting with $j^2$. The central idea is to show that every row starting with the sixth contains at least one element that occurs in a previous row. But this is true; we consider the first entry of a row if it is composite, the second otherwise. The rest is just strong induction on the rows. The final answer should be $\boxed{\mathbb Z^{+}\setminus \{1,2\}}$

Can you please show what $\boxed{\mathbb Z^{+}\setminus \{1,2\}}$ means for the first several terms? Thanks

All positive integers but $1$ and $2$.

I suppose I should be more explicit about the solution. Again we will construct the triangle. An image of the first sixteen rows is attached below. The motivation behind the construction is that two entries in the same row must be "guererra." Specifically, the $i^{\text{th}}$ row is the set of $n\in\mathbb Z^{+}$ such that there exist $a,b\in\mathbb Z^{+}$ satisfying $a+b=i+1$. Note that one integer can show up in many places on the triangle (in fact this is the crux of the solution). What our operation does is paint every unpainted row containing a integer that is (somehwere else) painted. It is easy to check that the third, fourth, and fifth rows are painted after a few iterations of the operation. Now, consider the $i^{\text{th}}$ row where $i\geq 6$. If the first entry, which is $i$, is composite, then there is an "earlier" row containing $i$. This is because if we write $i=ab$ with $a,b>1$ then $i=1+(i-1)>a+b$. Otherwise if $i$ is prime then $i-1$ is not prime, and then the second entry, $2(i-1)$ can be factored as $ab$ with $a,b>2$ whenever $i\geq 7$, and again we can get $i=2+(i-2)>a+b$. Now suppose all the rows less than $i$ are painted (excluding the first and the second). Then some element of row $i$ must be painted, and hence row $i$ will also get painted. In this manner we obtain the answer $\boxed{\mathbb Z^{+}\setminus \{1,2\}}$ Edit: I know that my typing is not so good; please ask questions if anything is unclear.

Attachments:

The main idea is to note $2ab=(2a)b=(2b)a$. Thus $2a+b-1=(2a+b-1)1$ is guerrera with $2ab$ which is also guerrera with $2b+a-1$. Thus for positive integers $a,b$, $2a+b-1$ is coloured iff $2b+a-1$ is. Substitute $b=a+1$; then $2a+(a+1)-1=3a$ is coloured iff $2(a+1)+a-1=3a+1$ is too. Subs $b=a+2$; $2a+(a+2)-1=3a+3$ iff $2a+(a+2)-1=3a+1$. Given that $3$ is coloured, we get that all $3a,3a+1$ ($a$ positive) is coloured. Substitute $b=a+3$; then $2a+(a+3)-1=3a+2$ is coloured iff $2(a+3)+a-1=3a+5$. Then as $5$ is coloured, all $3a+2$ ($a$ positive) is coloured. 2,1 are uncoloured as there are not guerrera with any other number. So all $\mathbb Z^{+} \setminus \{1,2\}$ is coloured.