imc

Since $x,y,z$ are all reals located in $[0,n]$, the number of choices for each one is continuous so we use geometric probability. WLOG(Without loss of generality), assume that $n\ge x\ge y\ge z\ge 0$. Then the set of points $(x,y,z)$ is a tetrahedron, or a triangular pyramid. The point $(x,y,z)$ distributes uniformly in this region. If this is not easy to understand, read Solution II. The altitude of the tetrahedron is $n$ and the base is an isosceles right triangle with a leg length $n$. The volume is $V_1=\frac{n^3}{6}$, as shown in the first figure in red. Now we will find the region with points satisfying $|x-y|\ge 1$, $|y-z|\ge 1$, $|z-x|\ge 1$. Since $n\ge x\ge y\ge z\ge 0$, we have $x-y\ge 1$, $y-z\ge 1$. The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too. The volume of this region is $V_2=\frac{(n-2{)}^3}{6}$. So the probability is $p=\frac{V_2}{V_1}=\frac{(n-2{)}^3}{n^3}$. Substituting $n$ with the values in the choices, we find that when $n=10$, $p=\frac{512}{1000}>\frac{1}{2}$, when $n=9$, $p=\frac{343}{729}<\frac{1}{2}$. So $n\ge 10$. So the answer is $\boxed{10}$.