Let $p\in(0,1)$, $n$ a positive even integer, $k,l\in\{0,\dots,n\}$, and $X\_k\sim \text{Binomial}(k,p)$, $Y\_{n-k}\sim \text{Binomial}(n-k,1-p)$ independent random variables. I would like to prove that
$$
\Pr(X\_k+Y\_{n-k}=l)\leq\Pr(X\_{n/2}+Y\_{n/2}=n/2).
$$


This question can be stated analytically. Setting $c=(1-p)/p$, define:
$$
f\_{n,c}(k,l)=c^{l+k}\sum\_{i=\max(0,k+l-n)}^{\min(k,l)}\binom{k}{i}\binom{n-k}{l-i}c^{-2i}.
$$
Prove that $f\_{n,c}$ attains its maximum at $k=l=n/2$, for any even $n$ and $c>0$.



A sum of two binomial random variables
Here is a (surprising) proof using Cauchy-Schwarz and "rearrangement".
The following lemma will be the key.


**Lemma**
: Let $X,Y$ be independent integer-valued rvs, then \begin{align\*}
(a)\; &\mbox{ for any } z: \;\mathbb{P}(X+Y=z)^2\leq \big(\sum\_x\mathbb{P}(X=x)^2\big)\,\big(\sum\_y \mathbb{P}(Y=y)^2\big)\\
(b)\; &\sum\_z\mathbb{P}(X-Y=z)^2=\sum\_z\mathbb{P}(X+Y=z)^2\end{align\*}
**Proof**: (a) apply Cauchy-Schwarz to
$\mathbb{P}(X+Y=z)=\sum\_x \mathbb{P}(X=x)\mathbb{P}(Y=z-x)$


(b) let $(X^\prime,Y^\prime)$ be distributed as $(X,Y)$, and independent of $(X,Y)$. Then
\begin{align\*}
 \sum\_z\mathbb{P}(X-Y=z)^2=\mathbb{P}(X-Y=X^\prime-Y^\prime)=\mathbb{P}(X-X^\prime=Y-Y^\prime)\\
 \sum\_z\mathbb{P}(X+Y=z)^2=\mathbb{P}(X+Y=X^\prime+Y^\prime)=\mathbb{P}(X-X^\prime=Y^\prime-Y)
\end{align\*}
Since $Y-Y^\prime$ and $Y^\prime-Y$ are identically distributed, and independent of $(X,X^\prime)$ the right hand sides above are equal. **End Proof**


Now to your question above. Let $n=2m$. We have to show that 
\begin{align\*} \mathbb{P}(X\_k+Y\_{2m-k}=\ell)\leq \mathbb{P}(X\_m+Y\_m=m)\end{align\*}For the right hand side above we have (using 1. below)
\begin{align\*} \mathbb{P}(X\_m+Y\_m=m)=\mathbb{P}(X\_m=X\_m^\prime)=\sum\_l \mathbb{P}(X\_m=l)^2\end{align\*}
To transform the left hand side, observe that well known properties of the binomial distribution give:
\begin{align} 1.\; &Y\_k \mbox{ is distributed as}\; k-X^\prime\_k, \mbox{where $X\_k^\prime$ is distributed as $X\_k$,}\\&\mbox{ and independent of $X\_k$}\\
2.\; &\mbox{ $X\_{m+j}$ resp. $Y\_{m+j}$ are distributed as $X\_m+X\_j$ resp. $Y\_m+Y\_j$, where}\\
&\mbox{ the summands are independent}
\end{align}
Using 1. gives that $\sum\_l \mathbb{P}(X\_m=l)^2=\sum\_l \mathbb{P}(Y\_m=l)^2$ and further that
$$ \mathbb{P}(X\_k+Y\_{2m-k}=\ell)=\mathbb{P}(X\_{2m-k}+Y\_k=2m-\ell)\;,$$ so we may w.l.o.g. assume that $k\leq m$.
Using 1. and 2. we have 
$$\mathbb{P}(X\_k+Y\_{2m-k}=\ell)=\mathbb{P}(X\_k-X\_{m-k}+Y\_m=\ell+k-m)$$
where $X\_k,X\_{m-k}$ and $Y\_m$ on the right hand side are independent. 


Using part (a) of the lemma (with $X=X\_{k}-X\_{m-k}$ and $Y=Y\_m$)
gives $$\mathbb{P}(X\_k+Y\_{2m-k}=z)^2 \leq \big(\sum\_x \mathbb{P}(X\_k-X\_{m-k}=x)^2\big)\big(\sum\_y\mathbb{P}(Y\_m=y)^2\big)$$
Finally using part (b) of the lemma on the first factor gives
$$\sum\_{x}\mathbb{P}(X\_k-X\_{m-k}=x)^2=\sum\_x\mathbb{P}(X\_k+X\_{m-k}=x)^2=\sum\_x\mathbb{P}(X\_m=x)^2$$
so that ultimately
$$\mathbb{P}(X\_k+Y\_{n-k}=z)^2\leq \big(\sum\_x \mathbb{P}(X\_m=x)^2\big)\big(\sum\_y\mathbb{P}(Y\_m=y)^2\big)=\big(\sum\_x \mathbb{P}(X\_m=x)^2\big)^2\;,$$
as desired.


