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
$$\sum_{k=1}^n\binom{n+k-1}{k-1}\binom{2n}{n+k}=
\sum_{k=0}^{n-1}\binom{2n}k2^k(-1)^{n-1-k}.$$