Let $A \subset [n]$ denote the subset of indices with $|x_i - y_i| \leq 1$, and $B$ its complement in $[n]$. It suffices to bound
$$\displaystyle \prod_{i \in B} |x_i - y_i|.$$
Let $t = \min_{i \in B} |x_i - y_i|$. By assumption, we have $t > 1$. Now, for any $s > 1$ I claim that the set of indices $B_s$ in $[n]$ with $|x_i - y_i| \geq s$ has at most $n/s$ elements. Indeed, by non-negativity it follows that for each $i \in B_s$ we have $\max\{x_i, y_i\} \geq s$. By monotonicity, it follows that if $x_i \geq s$ (say) then $x_j \geq s$ for $j \leq i$. Thus it follows that if $k$ is the largest index such that $x_k \geq s$ then
$$\displaystyle n \geq x_1 + \cdots + x_k \geq sk.$$
It thus follows that $k \leq n/s$, as desired.
Now set $s = t$. By construction we have
$$\displaystyle \prod_{i = 1}^n |x_i - y_i| \leq \prod_{i \in B} |x_i - y_i| \leq t^{n/t}.$$
The function on the right hand side is easily seen to have a global maximum at $t = e$, and hence the right hand side is bounded by $e^{n/e} < e^{n/2}$.