Proof by contradiction: assume there are no such pairs.

We can easily see that there must be infinitely many numbers of both colors, e.g. if there were only finitely many black numbers then any two numbers greater than the largest black number would both be white, as well as their sum.

Let a, b, c be three distinct numbers of the same color. Without loss of generality, let them be black. We'll add/subtract pairs of numbers of the same color to get new numbers that must be the opposite color to satisfy our assumption.

a + b : white
b + c : white
(a + b) + (b + c) = a + 2b + c : black
(a + 2b + c) - c = a + 2b : white
(a + 2b + c) - a = 2b + c : white
(a + 2b) + (2b + c) = a + 4b + c : black
(a + 4b + c) - (a + 2b + c) = 2b : white

Since there's nothing special about b (we can always find two other numbers of the same color as any given number) this means that for any number x, 2x must be the opposite color. So 4b must be black.

4b - b = 3b : white
But now we have a contradiction: a and 4b are both black, while 2b and 3b are both white. So a + 4b = 2b + 3b = 5b must be both black and white, which is impossible.

Proof by contradiction: assume there are no such pairs.

We can easily see that there must be infinitely many numbers of both colors, e.g. if there were only finitely many black numbers then any two numbers greater than the largest black number would both be white, as well as their sum.

Let a, b, c be three distinct numbers of the same color. Without loss of generality, let them be black. We'll add/subtract pairs of numbers of the same color to get new numbers that must be the opposite color to satisfy our assumption.

a + b : white
b + c : white
(a + b) + (b + c) = a + 2b + c : black
(a + 2b + c) - c = a + 2b : white
(a + 2b + c) - a = 2b + c : white
(a + 2b) + (2b + c) = a + 4b + c : black
(a + 4b + c) - (a + 2b + c) = 2b : white

Since there's nothing special about b (we can always find two other numbers of the same color as any given number) this means that for any number x, 2x must be the opposite color. So 4b must also be black.

4b - b = 3b : white
But now we have a contradiction: b and 4b are both black, while 2b and 3b are both white. So b + 4b = 2b + 3b = 5b must be both black and white, which is impossible.

Proof by contradiction: assume there are no such pairs.

We can easily see that there must be infinitely many numbers of both colors, e.g. if there were only finitely many black numbers then any two numbers greater than the largest black number would both be white, as well as their sum.

Let a, b, c be three distinct numbers of the same color. Without loss of generality, let them be black. We'll add/subtract pairs of numbers of the same color to get new numbers that must be the opposite color to satisfy our assumption.

a + b : white

b + c : white

 (a + b) + (b + c) = a + 2b + c : black

 (a + 2b + c) - c = a + 2b : white

 (a + 2b + c) - a = 2b + c : white

 (a + 2b) + (2b + c) = a + 4b + c : black

 (a + 4b + c) - (a + 2b + c) = 2b : white

Since there's nothing special about b (we can always find two other numbers of the same color as any given number) this means that for any number x, 2x must be the opposite color. So 4b must be black.

4b: black

4b - b = 3b : white

But now we have a contradiction: a and 4b are both black, while 2b and 3b are both white. So a + 4b = 2b + 3b = 5b must be both black and white, which is impossible.

Proof by contradiction: assume there are no such pairs.

We can easily see that there must be infinitely many numbers of both colors, e.g. if there were only finitely many black numbers then any two numbers greater than the largest black number would both be white, as well as their sum.

Let a, b, c be three distinct numbers of the same color. Without loss of generality, let them be black. We'll add/subtract pairs of numbers of the same color to get new numbers that must be the opposite color to satisfy our assumption.

a + b : white
b + c : white 
(a + b) + (b + c) = a + 2b + c : black 
(a + 2b + c) - c = a + 2b : white 
(a + 2b + c) - a = 2b + c : white 
(a + 2b) + (2b + c) = a + 4b + c : black 
(a + 4b + c) - (a + 2b + c) = 2b : white

Since there's nothing special about b (we can always find two other numbers of the same color as any given number) this means that for any number x, 2x must be the opposite color. So 4b must be black.

4b - b = 3b : white
But now we have a contradiction: a and 4b are both black, while 2b and 3b are both white. So a + 4b = 2b + 3b = 5b must be both black and white, which is impossible.