Just for the record, I noticed that if we take a common denominator for $a_n$ as coming out of the terms $n+k$ then one would get $\prod_{j=1}^n(n+j)=\frac{(2n)!}{n!}$. Hence, we may rewrite the given sequence as follows: \begin{align*} \sum_{k=1}^n\binom{n}k\frac{k}{n+k} &=\frac{n!}{(2n)!}\sum_{k=1}^nk\binom{n}k\frac{\prod_{j=1}^n(n+j)}{n+k} \\ &=\binom{2n}n^{-1}\sum_{k=1}^n\frac{(n+1)\cdots(n+k-1)}{(k-1)!}\cdot\frac{(n+k+1)\cdots(2n)}{(n-k)!} \\ &=\binom{2n}n^{-1}\sum_{k=1}^n\binom{n+k-1}{k-1}\binom{2n}{n+k} \end{align*} and the numerators (evidently integers) actually agree with what Max Alekseyev's referral to OEIS A240721. As an aside, we gather the identity that $$\sum_{k=1}^n\binom{n+k-1}{k-1}\binom{2n}{n+k}= \sum_{k=0}^{n-1}\binom{2n}k2^k(-1)^{n+k-1}.$$$$\sum_{k=1}^n\binom{n+k-1}{k-1}\binom{2n}{n+k}= \sum_{k=0}^{n-1}\binom{2n}k2^k(-1)^{n-1-k}.$$
