Answer: in red color: $1, 3, 5, ..., 2k+1$ for some $k$ (maybe, no one number is not red, or all odd numbers are red), and others numbers are blue Let's first note that all powers of two greater than 1 are blue. Really, $1+1, 2+2,..., 2^ n+2^n$ are always blue. Further, the sum of any several even powers of two is also blue, which can be proved, say, by induction. Now note that any even number is represented by the sum of various even powers of two. Then we get that all even numbers are colored blue. Then all the coloring pages turn out like this: all even numbers are blue, and odd numbers are colored like this: the first few consecutive odd numbers are colored red, the rest are blue. Firstly, they are all suitable, since the sum of two blue ones of different parity is greater than any red one, hence the blue number, in addition, the sum of numbers of the same parity is a blue number (because it is even). Secondly, if $b$ is a blue number, then $b+2, (b+2)+2,...., b+2n$ are also blue (by induction, say). It follows that all odd numbers in red must be $1, 2,...,2k+1$.

kraDracsO wrote: A coloration of the set of integer greater or equal to 1, must follow the next rule: each number is colored blue or red, so that the sum of two numbers (not necessarily different) of the same color be blue. Determine all the possible colorations of the set of the integers greater or equal to 1, follow this rule. $\color{blue}\boxed{\textbf{Answer: All colorings have the following form: the first } t \textbf{ odd numbers are red, and all other positive integers are blue}, \forall t=0,1,2,3,...,\infty}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{Claim 1:}}$ $\color{red}\rule{24cm}{0.3pt}$ $$\color{red}\textbf{All even numbers are blue}$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{Proof:}}$ $\color{red}\rule{24cm}{0.3pt}$ Let $n\in\mathbb{Z}^+$ Obviously $n$ has the same color as $n:$ $$\Rightarrow n+n=2n \text{ is blue}$$$$\Rightarrow \text{All even numbers are blue}_\blacksquare$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{Claim 2:}}$ $\color{red}\rule{24cm}{0.3pt}$ $$\textbf{If there is an odd blue number }m\textbf{ then all numbers }\ge m\textbf{ will be blue}$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{Proof:}}$ $\color{red}\rule{24cm}{0.3pt}$ By $\textbf{Claim 1}:$ $$2 \text{ is blue}$$Since $m$ is blue: $$\Rightarrow m+2 \text{ is blue}$$$$\Rightarrow m+4\text{ is blue}$$$$\Rightarrow m+6\text{ is blue}$$$$\vdots$$$$\Rightarrow m+2k\text{ is blue},\forall k\in\mathbb{Z}^+_0$$$$\text{All odd }\ge m\text{ are blue, and together with }\textbf{Claim 1}\text{ we have }\textbf{Claim 2}_\blacksquare$$$\color{red}\rule{24cm}{0.3pt}$ We put together what was obtained in $\textbf{Claim 1}$ and $\textbf{Claim 2}$, and we obtain: $$\boxed{\textbf{All colorings have the following form: the first } t \textbf{ odd numbers are red, and all other positive integers are blue}, \forall t=0,1,2,3,...,\infty}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

hectorleo123 wrote: $$\color{red}\textbf{All even numbers are blue}$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{Proof:}}$ $\color{red}\rule{24cm}{0.3pt}$ Let $n\in\mathbb{Z}^+$ Obviously $n$ has the same color as $n:$ $$\Rightarrow n+n=2n \text{ is blue}$$$$\Rightarrow \text{All even numbers are blue}_\blacksquare$$ Clever, more easier than my

This problem was proposed by Cristhian González and myself Santiago Rodriguez both of us from Colombia. I hope that you enjoy our problem.

Lemma 1: any even integer is blue Proof: $a+a=2a$ must be blue $\hspace{18cm}\blacksquare$ Lemma 1: If $2k+1$ is the first blue odd integers then, all the following odd integers are also blue Proof: The following integers are blue $$(2k+1)+2=2k+3$$$$(2k+3)+2=2k+5$$$$(2k+5)+2=2k+7$$\[\hspace*{3.3em} \vdots \]$\hspace{18cm}\blacksquare$ Therefore, the first $k$ odd numbers are red, and all the other numbers are blue, for all $k\geq 0$

First observe that an even natural number has the form $2n=n+n$ for some natural number $n$ and is therefore blue. If every odd integer is red then we have a valid colouring, else there exists a smallest blue odd natural number $n$, so if $m>n$ is an odd natural number then $m=n+(m-n)$, with $m-n\in \mathbb{N}$ even and therefore blue, hence $m$ is blue. In conclusion, the valid colourings are the ones with either all or the first $k$ odd numbers red for some $k\in \mathbb{N}_0$ and the remaining numbers blue.

All even numbers are blue since any integer has the same color has itself. Then if $n$ is the smallest blue odd number, $n$ and $2k$ are both blue for any positive integer $k$, so all odd numbers greater than $n$ are also blue. Thus, the only red numbers are the first $x$ odd natural numbers for some nonnegative integer $x$ (its easy to see that this works).

can we say sum of other colours number mustn't be blue

I went to this OMCC and felt this problem very easy. The solution is as follows. Consider some positive integer $n$. Obviously, $n$ has the same color as itself, then $n + n = 2n$ is blue. Therefore, all even numbers are blue. Now, consider the smallest odd positive integer $2k+1$ that is blue. Then, as we proved, 2 is blue, so $2k+3$ is also blue. By the same reasoning, $2k+5,2k+7, \dots$ are all blue, so therefore all odd numbers $\geq$ than $2k+1$ are blue. If there's no minimum odd positive integer that is blue, the all odd natural numbers are red. Therefore we get the following 3 solutions: 1. All odd positive integers are red and even positive integers are blue. 2. All natural numbers are blue. 3. The first $k$ odd numbers $1,3,5, \dots ,2k-1$ are red and the other natural numbers are blue. All of these clearly work.