Let $c=2000(2 d+2)$. Then from the equality $c=2000((d+1)+(d+1))$, it follows that the numbers $d+1$ and $c$ are of the same color.

On the other hand, $c=2000(d+(d+2))$, which means that the numbers $d, d+2$ and $d+1$ are either of the same color or of three different colors.

Therefore, any three consecutive numbers are either of the same color or of three different colors. If the numbers 1, 2, 3 are of the same color, then considering the triples $2,3,4 ; 3,4,5$ and so on, we get that all numbers are of the same color.

If 1 is of color $A$, 2 is of color $B$, and 3 is of color $C$, then from the triple 2, 3, 4 we get that 4 is of color $A$; from the triple $3,4,5$: that 5 is of color $B$, and so on.

Let $a=3 k_{1}+r_{1}, b=3 k_{2}+r_{2}, c=3 k_{3}+r_{3}$ ( $r_{1}, r_{2}, r_{3}$ - the remainders of the numbers $a, b, c$ when divided by 3 ).

The equality $2000(a+b)=2000\left(3 k_{1}+r_{1}+3 k_{2}+r_{2}\right)=3 M-\left(r_{1}+r_{2}\right)=c=3 k_{3}+r_{3}$ is possible only when
$r_{1}+r_{2}+r_{3}$ is divisible by 3, that is, either when the remainders $r_{1}, r_{2}, r_{3}$ are equal, or when they are pairwise distinct.

From this it follows that the found colorings satisfy the condition.

## Answer

Two colorings:

all numbers of one color

numbers $3 k-2, k \in \mathbf{N}$ - color $A$, numbers $3 k-1$ - color $B$, numbers $3 k$ - color $C$.