Integer roots of a symmetric polynomial

Fix large $N$ and consider all $n$-tuples $(x_1,\dots,x_n)\in \{1,\dots,N\}^n$. There are $N^n$ such $n$-tuples, at least $N^n/n!$ tuples modulo permutations, and for them the pairs $(x_1+\dots+x_n,x_1^2+\dots+x_n^2)$ take at most $n\cdot N\cdot n\cdot N^2=n^2N^3$ possible values. Thus by pigeonhole principle some value is obtained at least $N^{n-3}/(n^2\cdot n!)$ times. This is greater than 1 if $n>3$ and $N$ is chosen large enough.

Yes, there are. For instance, (1,4,6,7,2,3,5,8)

The general principle behind this solution is that $$ n^2+(n+1)^2=((n-1)^2+(n+2)^2)-4. $$ Combining two such collections on the opposite sides always gives you a solution.