Slightly stronger, there must be $2000$ zeroes in $(0,1)$. Suppose the contrary, that $f$ has fewer than $2000$ zeroes in $(0,1)$, then $f$ has fewer than $2000$ sign changes in $(0,1)$; let $x_0,...,x_n$ be the points at which $f$ changes signs with $n<2000$. Then $f(x)\prod_{k=1}^n (x-x_k)$ has no sign changes on $(0,1)$, and therefore $\left \vert \int_0^1 f(x)\prod_{k=1}^n (x-x_k)dx \right \vert > 0$. This is a contradiction, since $\int_0^1 f(x)P(x)dx=0$ for all polynomials $P$ of degree $1999$ or less.