Consider the following property : ( * ) if $n\geq 1$, then $a_n=\binom{2n}{n}$ is divisible by $2n-1$. One can show that ( * ) is true as follows : $2n-1$ divides $na_n$ (because of the identity $2(2n-1)a_{n-1}=na_n$) and $n$ and $2n-1$ are coprime, so that by Gauss' lemma $2n-1$ must divide $a_n$.
The natural reaction to this sort of proof is look for more "direct" proofs. For example, can $\frac{a_n}{2n-1}$ be written as an "obviously integral" expression (say, a product of sums of binomial coefficients or something similar).
I don't think there's an answer that truly satisfies your conditions, because we are working with binomial coefficients. For example, we know that $\binom{2n}{n}$ is integral, but can you write it as an obvious integral expression (combinatorial arguments excluded)?
Every proof will eventually boil down to the coprimality of $2n-1$ and $n$ I guess.