Alternative approach: Use Legendre's formula.
Given that $~n~$ is even, and that $~0 < k < n ~: ~k ~$ is odd.
To prove:
$$\sum_{i=1}^\infty \left\lfloor \frac{n}{2^i} \right\rfloor >
\left\{ ~\sum_{i=1}^\infty \left\lfloor \frac{k}{2^i} \right\rfloor ~\right\} + \left\{ ~\sum_{i=1}^\infty \left\lfloor \frac{n-k}{2^i} \right\rfloor ~\right\}.
\tag1 $$
(1) above can be proven column by column, be examining each value of $~i,~$ in turn.
$\underline{i = 1}$
Since $~n~$ is even, there exists a positive integer $~j~$ such that $~\displaystyle n = 2j \implies \left\lfloor \frac{n}{2^1} \right\rfloor = j.$
Since $~k~$ is odd, there exists a non-negative integer $~r~$ such that $~\displaystyle k = 2r+1 \implies \left\lfloor \frac{k}{2^1} \right\rfloor = r.$
Since $~n~$ is even and $~k~$ is odd, $~n-k~$ is odd.
Therefore, there exists a non-negative integer $~s~$ such that $~\displaystyle n-k = 2s+1 \implies \left\lfloor \frac{n-k}{2^1} \right\rfloor = s.$
Further, you have that $~2j = n = k + (n-k) = (2r + 1) + (2s+1) \implies ~$
$2j = 2(r + s + 1) \implies j > (r + s).$
Therefore, when evaluating the assertion in (1) above, you have that strict inequality holds when $~i = 1.~$
$\underline{i \geq 2}$
Therefore the entire problem is reduced to showing that for all $~i \in \Bbb{Z_{\geq 2}},~$ you have that
$$\left\lfloor \frac{n}{2^i} \right\rfloor \geq
\left\lfloor \frac{k}{2^i} \right\rfloor + \left\lfloor \frac{n-k}{2^i} \right\rfloor.
\tag2 $$
However, the assertion in (2) above is an immediate consequence of the (easily proven) lemma that given any $~x,y,z \in \Bbb{Z^+},~$ that
$$\left\lfloor \frac{x+y}{z} \right\rfloor \geq
\left\lfloor \frac{x}{z} \right\rfloor + \left\lfloor \frac{y}{z} \right\rfloor. \tag3 $$
That is, the assertion in (3) is easily proven by assuming that $~x,y~$ can each be expressed as
$~x = Mz + u ~: ~M \in \Bbb{Z_{\geq 0}}, ~u \in \{0,1,\cdots,z-1\},$
$~y = Pz + v ~: ~P \in \Bbb{Z_{\geq 0}}, ~v \in \{0,1,\cdots,z-1\},$
and seeing where this leads.