smc

We know $C_k^n=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, so let's rewrite the expression as $\left(\frac{n-2k-1}{k+1}\right)\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. Notice that $$n-2k-1=n-2(k+1)+1=(n+1)-2(k+1).$$ This allows us to rewrite the expression as $$\left(\frac{(n+1)-2(k+1)}{k+1}\right)\left(\genfrac{}{}{0ex}{}{n}{k}\right).$$ From here, we just have to do some algebra to get $$\left(\frac{(n+1)-2(k+1)}{k+1}\right)\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\left(\frac{n+1}{k+1}-2\right)\frac{n!}{k!(n-k)!}$$ $$=\frac{(n+1)!}{(k+1)!(n-k)!}-2\cdot \frac{n!}{k!(n-k)!}$$ $$=\left(\genfrac{}{}{0ex}{}{n+1}{k+1}\right)-2\left(\genfrac{}{}{0ex}{}{n}{k}\right),$$ so $\left(\frac{n-2k-1}{k+1}\right)C_k^n$ is an integer $\boxed{\text{for all }k\text{ and }n}$.