Who has got any solution

First, it's Colombia (COL), not ILL (which does not exist). From Viete's relation we have $ \sum_{i=1}^n r_i = -n$. Then, from Holder (or repeated applications of Cauchy-Schwartz) we have $ n^{16} = \left(\sum_{i=1}^n r_i^{16}\right)\cdot n^{15} \geq \left(\sum_{i=1}^n r_i\right)^{16} = n^{16}$, hence equality, hence all roots $ r_i$ are equal (to $ -1$). Therefore the polynomial is $ p(x) = (x+1)^n$.

$ \sum^{16}_{i = 1} r_i^{16}$ or $ \sum^{n}_{i = 1} r_i^{16}$ ???

The latter (the one I used). It was a typo by orl, the same as ILL instead of COL. I know because I own the 1989 booklet containing the IMO Longlist.

Solution: We repeatedly apply Cauchy-Schwartz to obtain $(\sum r_i^{16})(1+1+...+1)\ge (\sum r_i^{8})^2\ge n^2$. Since $n$ is positive ($\text{deg}(p)>0$), we have that $(\sum r_i^{16})\ge n$. Hence we have the equality case of Cauchy, so $\boxed{r_1=r_2=...=r_n=-1}$. Our $p(x)=(x+1)^n$. $\blacksquare$