THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

If $1$ is colored red, then for any green $a>1$, we have $1\cdot a=a$ must be colored red, a contradiction. Hence all numbers must be red in this case.

Consider now that $1$ is colored green.

If $2$ is red, then $1+2=3$ must be green, so $2+3=5$ must also be green, contradiction. So $2$ must be green.

If $3$ is red, then $2+3=5$ must be green, contradiction. So $3$ must be green. If $4$ is red, then $1+4=5$ must be green, contradiction. So $4$ must be green.

Since $5$ has a different color than $1,2,3,4$, all the numbers $1+5=6$, $2+5=7$, $3+5=8$, $4+5=9$ must be green, and $2\cdot 5=10$ must be red. The numbers $6,7,8,9$, respectively $10$ cannot be obtained as the product, respectively the sum, of $5$ and another number, so that this coloring satisfies the conditions in the statement of the problem.

Consequently, there are two possible colorings: all numbers colored red, or $5$ and $10$ colored red and the other numbers green.