I would like to prove the following conjecture for $n$ positive integer:
$$\sum_{k=\lfloor \frac{n+1}{2} \rfloor}^{n}{k{k-1 \choose \lfloor \frac{n+1}{2}\rfloor - 1}} = \Big\lceil \frac{n}{2} \Big\rceil{n+1 \choose \lfloor \frac{n}{2} \rfloor}$$
and in particular, if $n=2m$:
$$\sum_{k=m}^{2m}{k{k-1 \choose m-1}} = m{2m+1 \choose m}$$
The LHS counts the sum of the maximum elements of each subset of $[n]=\{1,\ldots,n\}$ with size $\lfloor (n+1)/2 \rfloor$. For example for $n=3$ there are three subsets $\{1,2\}, \{1,3\}, \{2,3\}$ and the sum of maximum values is $2+3+3=8$.
I have searched with Approach Zero but I didn't find the above equality.
Any hints?
As regards the case $n=2m$, $$\sum_{k=m}^{2m}k\binom{k-1}{m-1}=m\sum_{k=m}^{2m}\binom{k}{m} =m\binom{2m+1}{m+1}=m\binom{2m+1}{m}.$$ where we applied Hockey-stick identity.
The case $n=2m-1$ can be treated in a similar way: $$\sum_{k=m}^{2m-1}k\binom{k-1}{m-1}=m\sum_{k=m}^{2m-1}\binom{k}{m} =m\binom{2m}{m+1}=m\binom{2m}{m-1}.$$