Since $P(x)$ is real, it is sufficient to require that $P(i)=P(-1)=0$ Let's say $P(x)=\sum_{i=0}^6 a_ix^k$. Our conditions require that $$a_0+a_2+a_4+a_6=a_1+a_3+a_5$$$$a_0+a_4=a_2+a_6$$$$a_1+a_5=a_3$$ This is equivalent to $$a_3=a_1+a_5=a_2+a_6=a_0+a_4$$Note that for any value of $a_3$, there are $(a_3-1)$ ways to pick $(a_1,a_5)$, $(a_2,a_6)$, $(a_0,a_4)$ So our total number of polynomials is $$\sum_{k=1}^{1999} (k-1)^3$$$$\sum_{k=0}^{1998} k^3$$$$\frac{1998^2\cdot 1999^2}{4}$$