To solve the problem of coloring the set of integers greater than or equal to 1 according to the given rules, we need to carefully consider the constraints imposed by the problem. The key condition is that the sum of any two numbers of the same color must be blue.

First, let's start by analyzing the situation:

1. Blue Coloring Rule: If two integers $a$ and $b$ are both of the same color (and that color is not explicitly defined as red or blue in the statement), then their sum $a + b$ must result in a blue number.

2. Pair Considerations:
- If we consider coloring all even numbers blue, then for any two even numbers $2a$ and $2b$, the sum is $2a + 2b = 2(a+b)$, which is also even, hence blue.
- Also, for an even number $2a$ and an odd number $2b+1$, their sum is $2a + (2b+1) = 2(a + b) + 1$, which is odd, so it allows flexibility in coloring odd numbers.

3. Odd Numbers:
- If both numbers are odd, say $2a+1$ and $2b+1$, their sum is $2a + 1 + 2b + 1 = 2(a + b + 1)$, which is even, thus blue.

Given these considerations, the feasible solution respects the requirement that the sum of any two numbers of the same color must be blue:

- Conclusion: To satisfy the rules, the only possible coloring is when all even numbers are blue, thereby ensuring that the even numbers (which when added together remain even) satisfy the blue condition directly. Odd numbers can then freely be colored as required to maintain the overall constraint.

Thus, the possible coloring of integers greater than or equal to 1, following the given rule, is:
$$\textbf{All even numbers are blue}$$
$$\boxed{\textbf{All even numbers are blue}}$$

This conclusion ensures that under all pairings of integers with the same color, whether they are even or odd, the summation remains consistent with the rule provided: resulting always in an even (blue) number.