Solution:

Zero is a red number: indeed, if $0$ were white, since there exists a red number $x$, we would have that $0 + x = x$ is white by the second property, a contradiction.

One is a white number: indeed, if one were red, since there exists a white number $y$, we would have that $y \cdot 1 = y$ is red by the third property, a contradiction.

If there are no red numbers other than zero, the thesis is trivial. Otherwise, let $k$ be the smallest red number greater than zero. Then every number that is not a multiple of $k$ is white: indeed, if $n$ is not a multiple of $k$, $n$ can be written in the form $n = qk + r$ with $0 < r < k$. We use induction on $q$.

If $q = 0$, $n$ is white by hypothesis. Assuming the hypothesis holds for $q-1$, we have $n = [(q-1)k + r] + k$, which is white by the second property.

By the third property, every multiple of $k$ of the form $j \cdot k$, with $j$ not divisible by $k$, is red. Suppose now that $n$ is a multiple of $k$ of the form $j \cdot k$ with $j = l k$ divisible by $k$, that is, that $n$ is of the form $l \cdot k^{2}$. From the equality $k + l \cdot k^{2} = (1 + l k) \cdot k$ we obtain, by the second property, that in this case too $n$ must be red. Therefore the red numbers are all and only the multiples of $k$. In this case both the hypotheses of the problem and the thesis are trivially verified.