Given an integer $n$, let us call $I_{n}$ the set of 2000 consecutive numbers $\{n, n+1, \ldots, n+1999\}$ of length 2000, and let us call $R_{n}$ the number of red integers and $B_{n}$ the number of blue integers in $I_{n}$: observe that, since $R_{n}+B_{n}=2000$ is even, $D_{n}=R_{n}-B_{n}$ is also even. Observe also that, in passing from $I_{n}$ to $I_{n+1}$, the number of red or blue naturals can change by at most one unit, and therefore $D_{n}-2 \leq D_{n+1} \leq D_{n}+2$.

Suppose for contradiction that for no $n$ do we have $D_{n}=0$, and suppose, without loss of generality, that $D_{0}<0$. We want to show, by induction, that $D_{n}<0$ for every $n \geq 0$: the base step holds by hypothesis. $D_{n}$ is even and negative, and $D_{n+1} \leq D_{n}+2<2$ is even, and is not zero by hypothesis, and therefore $D_{n+1}$ is also negative.

Let us now consider the intervals $I_{2000 k}$ as $k$ ranges over the interval $0 \leq k \leq 500$. The $I_{2000 k}$ are pairwise disjoint, and in each of them there are at least 1001 blue numbers and at most 999 red numbers, hence in the interval $\{0,1, \ldots, 2000 \cdot 500+1999\}$ there are at least $1001 \cdot 501$ blue numbers and at most $999 \cdot 501$ red numbers, contradicting the hypothesis.