The identity $(n+1) \binom {2n}{n-1} = n \binom {2n}n$ shows that $(n+1)|n \binom {2n}n$ and, since $\gcd(n, n+1) = 1, (n+1)|\binom {2n}n.$

We know that $\frac 1{n+1} \binom{2n}n$ is the number of triangularisations of a $n$-gon (the $n$-th Catalan number). This number is trivially an integer.

Peter wrote: Prove that $ 2n\choose n$ is divisible by $ n+1$. One may slightly generize the argument of mathmanman. Proposition. Let $ N$ and $ k$ are positive integers with $ N\geq k$ and $ \gcd(N+1, k+1)=1$. Then, $ N\choose k$ is divisible by $ k+1$. Proof. It is well-known that \[ (N+1){N\choose k}= (k+1){{N+1}\choose{k+1}}.\] Hence, $ (N+1){N\choose k}$ is divisible by $ k+1$. Since $ \gcd(N+1, k+1)=1$, this implies that $ N\choose k$ is divisible by $ k+1$.

As you know,$C_n=\frac{1}{n+1}\binom{2n}{n}$ is called as Catalan number. $\binom{2n}{n}-\binom{2n}{n-1}=\frac{(2n)!}{n!n!}-\frac{(2n)!}{(n-1)!(n+1)!}$ $=\frac{(2n)!}{n!(n+1)!}=\frac{1}{n+1}\binom{2n}{n}$ is positive integer.The proof is completed.$\blacksquare$