Answer is $n(n-1)$. Value $n(n-1)$ attained when $x_1=0$ and $x_2=\dots=x_n=n$. If some $x_i=n$ then some $x_j=0$, so $y\leq n(n-1)$ since $x_i\in [0;n]$. Now suppose that all $x_i\in (0;n)$. Denote $y_i=\frac{x_i}{n-x_i}$. Then $x_i=n\cdot\frac{y_i}{y_i+1}$. We suffices to prove that $$ \sum_{i=1}^n \frac{y_i}{y_i+1}\leq n-1 $$if $\Pi_{i=1}^ny_i=1$. Let $z_i=1/y_i$. Prove by induction that $$ \sum_{i=1}^n \frac{1}{z_i+1}\leq n-1 $$if $\Pi_{i=1}^nz_i=1$. For $n=2$ it's equality: $\frac{1}{z+1}+\frac{1}{1/z+1}=1$. To prove induction step observe that for positive $x,y$ $$ \frac{1}{1+x}+\frac{1}{1+y}\leq \frac{1}{1+xy}+1, $$$$ \frac{2+x+y}{(1+x)(1+y)}\leq \frac{1}{1+xy}+1 $$$$ \frac{1-xy}{(1+x)(1+y)}\leq \frac{1}{1+xy} $$and $1-x^2y^2\leq 1\leq (1+x)(1+y)$