Is the language of words that are unbalanced in the first half context-free?
Perhaps it can be proved using Ogden's Lemma and its generalization by Bader and Moura, this is a rather informal sketch of the proof.
First restrict $L$ to strings of length $4n$ and apply to it the following homomorphism between $\Sigma = \{ 0,1 \}$ and $\Sigma' = \{ a, b, c\}$:
$h(11) \to a$
$h(00) \to b$
$h(01) \to c$
$h(10) \to c$
If $L$ is CF then also the new language $L'$ obtained is CF by closure properties.
Informally $L'$ contains an unbalanced number of $a$ and $b$ in the first half and the number/occurences of $c$ doesn't matter.
Further restrict $L'$ by intersecting it with the regular language $R = \{ a^* (c^* b^*)^*  \}$; let $L'' = R \cap L'$
For example the string
$a a c b | cccc \in L''$ corresponds to $11\;11\;10\;00\; |\; 10\;10\;10\;10 \in L$ ($|$ is used to mark the half of the string for better readability)
$a b c c | cccc \notin L''$ corresponds to $11\;00\;10\;10\;|\;10\;10\;10\;10 \notin L$
Suppose that $L''$ is CF, and $p$ is its pumping length. Build $w \in L''$ concatenating these four parts:

$(\;a^p\;)$ leading $a$'s

$(\;c^j\;)$ a sequence of $c$s, we'll fix $j$ below

$(\;c^{p} \;b\;)$ repeated $p + p!$


If $n$ is the constant of the Bader-Moura's condition, then we pick $j$ large enough to exclude all the symbols in part 1 and 3: $j \geq n^{p+(p+1)(p+p!)+1}$

$(c^k)$ where $k$ is large enough to be pumped exluding all  previous symbols: $k \geq n^{p + j + (p+1)(p+p!)+1}$

$w = a^{p} \; c^j \; (c^{p} \;b )^{p+p!} \; c^{k} $
Now we mark the first $a$ sequence as distinguished, the string $vx$ must contain $0 &lt; q \leq p$ distinguished positions ($\#a_{vx} = q$) by Ogden's lemma; $vx$ can also contain one $b$ (not more than one because the $b$s are spaced with more than $p$ symbols $c$) and $0 \leq r &lt; p$ symbols $c$ ($\#c_{vx} = r$).

if $vx$ is such that $\#a_{vx}=q$, $\#b_{vx}=0$, $\#c_{vx}=r$:

then we can pump $i = p! / q $ times and we obtain the same number of $a$s and $b$s; if after the pump some $b$s fall after the half of the string, we can pump the final $c^k$ independently from the rest of the string and we can &quot;push&quot; all $a$s and $b$s back in the first half (and $\#a_{w'} = \#b_{w'} = p + p!$), so the pumped string $w'$ is not in $L''$

if $vx$ is such that $\#a_{vx}=q$, $\#b_{vx}=1$, $\#c_{vx}=r$:

then each time we pump we increase the number of $a$s and $b$s, but we cannot guarantee that we reach the same number (e.g. in the case $\#a_{vx}=q=1\#b_{vx}$). But in this case the derivation tree &quot;isolate&quot; the $c^j$ part of the string from the final part $c^k$, so we can pump them independently.
We can pump $c^j$ as many time as needed to &quot;push&quot; $p!$ symbols $b$s to the second half of the string. Suppose that the pumping length of $c^j$ is $s$ (that must be even), the half of the string is shifted towards the $b$s by $s/2$. We have $s \leq p$ so after each pump at most one $b$ is &quot;pushed&quot; to the second half, because the &quot;distance&quot; between $b$s is $p$. So also in this case we get a string $w'$ not in $L''$