Any help with this alternate sum Conjecture: Given any positive integer $d\ge2$, there exits $k$ such that  $$\binom{2k}{k+d-j}-\binom{2k}{k-j}+\binom{2k}{k-d-j}-\binom{2k}{k-2d-j}>0,$$ for all $0\le j \le d-\lfloor d/2 \rfloor -1.$
PD: I have verified this for $2 \le d \le 300$.$\:$ Moreover, it seems that each of this sums tends to infinity as $k$ goes to infinity but I just need to show that these are positive for some $k$. I have asked a similar question before but the present one is less restrictive.
 A: Pre-note: This is not a rigorous answer, regard it as heuristic. It relies on an initial inequality not shown, and a use of the normal approximation to get the binomials.
Denote the alternating sum by $f(k,d,j).$ Experimentally it seems that for fixed $k,d$ we have as $j$ increases in the sequence $0,1,2,\cdots$ that $f$ also increases. (It may start out negative, but goes up through less negative and maybe into positive, at least as long as the "denominators" of the binomial coefficients all stay in range.) I certainly do not have a proof of this part, although in cases where a $k$ that "works" for a given $d$ and $j=0$ the inequality seems to hold without the limitation on $j$ being at most $d-[d]-1$ with $[x]$ meaning floor.
The distribution of the binomial coefficients can be approximated by a normal distribution. That is, we consider $X$ to be binomial with $2k$ repeats, with success probability $1/2$. Then $X$ has mean $k$ and variance $k/2$, so standard deviation $\sqrt{k/2}.$ The $z$-score for a given $x$ is then $z=(x-k)/(\sqrt{k/2}),$ and then this $z$ is approximately normal with mean $0$ stsndard deviation $1$. 
Now suppose a given $d$ is fixed, we want to see how large a $k$ should be taken to make $f(k,d,0)>0.$ On the normal curve, with its density $q(t)=(1/\sqrt{2\pi})e^{-t^2/2},$ we note that $q(-1)-q(0)+q(1)-q(2) \approx +0.031,$ so that we can be "heuristically sure" of getting a positive value for $f(k,d,0)$ if we take $k$ large enough that the $z$-score of $d$ is less than $1.$ This is the same as putting $d$ equal to the standard deviation $\sqrt{k/2}.$ of the untransformed $X$, which gives $d=\sqrt{k/2}$ or $k=2d^2.$
From what I have found, this choice $k=2d^2$ always works and then some, to give the desired inequality. For example $d=5$ suggests by this heuristic that we take $k=50.$ And in fact that works, but so do each of $41,42,...,49.$ $k=40$ doesn't work, i.e. $f(40,5,0)<0.$ 
A: Let $p=\lfloor{d/2}\rfloor$. Note that for $0\leq j \leq d-p-1$ 
\begin{align*}
&C(2N,N+d-j)-C(2N,N-j)+C(2N,N-j-d)-C(2N,N-j-2d)
\\
&\geq C(2N,N+d)-C(2N,N)+C(2N,N-2d+p+1)-C(2N,N-2d). 
\end{align*}
Now,
\begin{align*}
&C(2N,N+d)-C(2N,N)+C(2N,N-2d+p+1)-C(2N,N-2d)
\\
&= \left\{\frac{(2N)!}{(N+d)!(N-d)!}-\frac{(2N)!}{N!N!}+\frac{(2N)!}{(N-2d+p+1)!(N+2d-p-1)!}-\frac{(2N)!}{(N-2d)!(N+2d)!} \right\}.
\end{align*}
Let 
\begin{align*}
&\left\{\frac{1}{(N+d)!(N-d)!}-\frac{1}{N!N!}+\frac{1}{(N-2d+p+1)!(N+2d-p-1)!}-\frac{1}{(N-2d)!(N+2d)!} \right\}
\\
&=\frac{q(N)}{(N+2d-p-1)!(N+2d)!},
\end{align*}
where
\begin{align*}
q(N)&=\prod_{i=1}^{d}(N+d+i)\prod_{i=1}^{3d-p-1}(N-d+i)-\prod_{i=1}^{2d}(N+i)\prod_{i=1}^{2d-p-1}(N+i)
\\
&+\prod_{i=1}^{4d-p-1}(N-2d+p+1+i)-\prod_{i=1}^{4d-p-1}(N-2d+i).
\end{align*} 
Considering $q(N)$ as a polynomial in the variable $N$, we want to show that the leading term is positive. You can check that this leading coefficient is $$\frac{1}{2}( 8dp-2d^2-2p^2+8d-4p-2)$$ and is easy to verify that this is positive.  Thus, $q(N)$ goes to infinity as $N$ goes to infinity.
