For a cleverness free solution:
The number of positive roots counting multiplicity is $2$: the root $1$ is repeated twice.
You can use Descartes' rule of signs, which is a general method which can be useful sometimes.
Your equation (after multiplying by $x^2-1$) is
$ P(x) = x^{2000} - 1000x^{1001} + 1000x^{999} - 1 = 0$
which has $3$ sign changes.
So the number of positive roots is either $1$ or $3$ (counting multiple roots multiple times).
The derivative is P'(x) = 2000 x^{1999} - 1000\times 1001 x^{1000} + 1000 \times 999 x^{998}
Notice that $P(1) = 0$, P'(1) = 0.
Since $1$ is a root, and also of the derivative, the number of positive roots is $3$, counting $1$ at least two times.
But since we introduced an extraneous positive root by multiplying by $x^2 -1$, the number of positive roots of your original equation, counting multiplicity is $2$.
The second derivative
P''(x) = 2000\times 1999 x^{1998} - 1000\times1000\times1000 x^{999} + 1000 \times 999 \times 999
and P''(1) = 0. Thus the root $1$ is of multiplicity $2$.
This also applies to the equation $x^{2n} - n(x^{n+1} - x^{n-1}) - 1 = 0$ for $n \gt 1$.
See Also: Sturm's theorem.