So $(x+1)P(x)-x$ is a polynomial of degree $2n+1$ with roots at $0,1,2,\dots,2n$ and hence \[(x+1)P(x)-x=cx(x-1)(x-2)\dots (x-2n)\]for some constant $c$. Putting $x=-1$ we find that $c=-\frac{1}{(2n+1)!}$. Putting $x=2n+1$ we find that $(2n+2)P(2n+1)-(2n+1)=-1$ and hence $\boxed{P(2n+1)=\frac{n}{n+1}}$.

parmenides51 wrote: Let $P (x)$ be a polynomial of degree $2n$, such that $P (k) =\frac{k}{k + 1}$ for $k = 0,...,2n$. Determine $P (2n + 1)$. This was Problem 3 in the USAMO 1975. See here.