<s>[INST] Looking for a combinatorial proof for a Catalan identity Let $C_n=\frac1{n+1}\binom{2n}n$ be the familiar Catalan numbers.

QUESTION. Is there a combinatorial or conceptual justification for this identity?
$$\sum_{k=1}^n\left[\frac{k}n\binom{2n}{n-k}\right]^2=C_{2n-1}.$$

 [/INST] Expanding my previous comment into an answer at the OP's request. We can write
$$
\frac{k}{n}\binom{2n}{n-k}=A_k-B_k,
$$
where
$$
A_k=\binom{2n-1}{n-k}=\binom{2n-1}{n+k-1}, \qquad B_k=\binom{2n-1}{n+k}=\binom{2n-1}{n-k-1}.
$$
Then
$$
\sum_{k=1}^{n}\left(\frac{k}{n}\binom{2n}{n-k}\right)^2=\sum_{k=1}^{n}(A_k^2+B_k^2)-\sum_{k=1}^{n}(A_kB_k+B_kA_k).
$$
The first sum on the right is
$$
\sum_{k=1}^{n}\left(\binom{2n-1}{n-k}\binom{2n-1}{n+k-1}+\binom{2n-1}{n+k}\binom{2n-1}{n-k-1}\right)=\\
=\left(\sum_{k=0}^{2n-1}\binom{2n-1}{k}\binom{2n-1}{2n-1-k}\right)-\binom{2n-1}{n}\binom{2n-1}{n-1}=\binom{4n-2}{2n-1}-\binom{2n-1}{n}^2,
$$
and the second sum on the right is
$$
\sum_{k=1}^{n}\left(\binom{2n-1}{n-k}\binom{2n-1}{n+k}+\binom{2n-1}{n+k}\binom{2n-1}{n-k}\right)=\\
=\left(\sum_{k=0}^{2n-1}\binom{2n-1}{k}\binom{2n-1}{2n-k}\right)-\binom{2n-1}{n}^2=\binom{4n-2}{2n}-\binom{2n-1}{n}^2,
$$
so
$$
\sum_{k=1}^{n}\left(\frac{k}{n}\binom{2n}{n-k}\right)^2=\binom{4n-2}{2n-1}-\binom{4n-2}{2n}=C_{2n-1}.
$$
</s>