Iranian Mathematical Olympiad

First suppose the case that none of the numbers are zero. Note that there exist at least 1000 positive numbers or at least 1000 negative numbers among these 2000 numbers. If there exist at least 1000 negative numbers and we put these 1000 numbers as roots of a degree 1000 polynomial all its coefficients are positive. So in every case we have 1000 positive numbers.

Now take 1000 positive numbers and put the 1000 remaining numbers as roots of a polynomial, all of its coefficients are positive, so the numbers must be negative. Therefore there is 1000 positive and 1000 negative numbers. If we put 1000 positive numbers as roots of a polynomial its coefficients are alternating positive and negative. This is a contradiction because remaining 1000 numbers all are negative. The contradiction shows that there at least one number equal to zero among numbers.

Denote by $k$ the number of zeros among numbers so $k > 0$. If $k < 1000$, then put these $k$ numbers zero and arbitrary $1000 - k$ numbers among others as roots of a polynomial. The product of roots is zero so there exists another number equal to zero and this contradicts the definition of $k$, thereby $k > 1000$. Now put 1000 numbers equal to zero as roots of a polynomial this polynomial is $x^{1000}$, so other numbers are equal to zero. So there do not exist 2000 numbers with mentioned property. □