Start by encoding the sum call it $S_n$ using residues. We have by
inspection that
$$S_n =
\frac{1}{2}\sum_{1\ \leq\ j,\ j'\ \leq\ n \atop j\ne j'}\
\prod_{k\ \neq\ j,\,j'}^{n}
{\left(\, j + j'\,\right)^{2} \over \left(\, j - k\,\right)\left(\, j' - k\,\right)}
= \frac{1}{2}
\sum_{q=1}^n
\mathrm{Res}
\left(f(z); z=q\right)$$
where
$$f(z) = \prod_{p=1}^n\frac{1}{z-p}
\sum_{j=1}^n (z-j)\times
\frac{(-1)^{n-1-j} (z-j)(z+j)^{2n-4}}{(j-1)!(n-j)!}.$$
This is because
$$\prod_{k\ne j, j'}^n \frac{(j+j')^2}{(j-k)(j'-k)}
= (j+j')^{2n-4}
\prod_{k\ne j, j'}^n \frac{1}{j-k}
\prod_{k\ne j, j'}^n \frac{1}{j'-k}$$
and we get from the residue at $z=j'$ from $f(z)$ for $j\ne j'$ the
contribution
$$(j'-j) \prod_{p\ne j'}^n \frac{1}{j'-p}
\times (j'-j) (-1)^{n-1-j} \frac{1}{(j-1)! (n-j)!}
\times (j+j')^{2n-4}$$
and we have
$$(j'-j) \prod_{p\ne j'}^n \frac{1}{j'-p} =
\prod_{p\ne j,j'}^n \frac{1}{j'-p}$$
and
$$(j'-j) (-1)^{n-1-j} \frac{1}{(j-1)! (n-j)!} =
\prod_{p\ne j,j'}^n \frac{1}{j-p}.$$
There is a zero contribution when $j=j'$ because the factor $(z-j)^2$
cancels the pole.
We can therefore collect $S_n$ by integrating $f(z)$ around a contour that encloses the $n$ poles.
We will then use the residue at infinity to evaluate the sum of the residues inside the contour.
Simplify $f(z)$ to obtain
$$f(z) = \frac{1}{(n-1)!}
\prod_{p=1}^n\frac{1}{z-p}
\sum_{j=1}^n {n-1\choose j-1}
(-1)^{n-1-j} (z-j)^2(z+j)^{2n-4}$$
which is
$$f(z) = \frac{1}{(n-1)!}
\prod_{p=1}^n\frac{1}{z-p}
\sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-j} (z-j-1)^2(z+j+1)^{2n-4}.$$
Now the residue at infinity of $f(z)$ is given by
$$\mathrm{Res}
\left(-\frac{1}{z^2} f\left(\frac{1}{z}\right); z=0\right).$$
The functional term becomes
$$-\frac{1}{z^2} \frac{1}{(n-1)!}
\prod_{p=1}^n\frac{1}{1/z-p} \\ \times
\sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-j} (1/z-j-1)^2(1/z+j+1)^{2n-4}.$$
This is
$$-\frac{1}{z^2} \frac{1}{(n-1)!}
\prod_{p=1}^n\frac{z}{1-pz} \\ \times
\sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-j} \frac{(1-(j+1)z)^2}{z^2}
\frac{(1+(j+1)z)^{2n-4}}{z^{2n-4}}$$
which becomes
$$-\frac{z^n}{z^2\times z^2\times z^{2n-4}}
\frac{1}{(n-1)!}
\prod_{p=1}^n\frac{1}{1-pz} \\ \times
\sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-j} (1-(j+1)z)^2
(1+(j+1)z)^{2n-4}.$$
Drop the minus sign (residues sum to zero) to obtain
$$S_n =
\frac{1}{2} \mathrm{Res}
\left(\frac{1}{z^n}
\frac{1}{(n-1)!}
\prod_{p=1}^n\frac{1}{1-pz} \\ \times
\sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-j} (1-(j+1)z)^2
(1+(j+1)z)^{2n-4}; z=0\right).$$
Using $$(1-(j+1)z)^2 = 1 - 2(j+1)z + (j+1)^2z^2$$ we get three pieces
from this, call them $A_n, B_n$ and $C_n.$ We will evaluate these making use of the disappearance of terms from formal power series in residue calculations and the fact that Stirling numbers of the second kind vanish when partitioning into more subsets than there are elements.
We have
$$A_n =
\frac{1}{2} \mathrm{Res}
\left(\frac{1}{z^n}
\frac{1}{(n-1)!}
\prod_{p=1}^n\frac{1}{1-pz} \\ \times
\sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-j} \sum_{q=0}^{2n-4} {2n-4\choose q} (j+1)^q z^q; z=0\right).$$
This becomes
$$A_n =
\frac{1}{2} \mathrm{Res}
\left(\frac{1}{z^n}
\frac{1}{(n-1)!}
\prod_{p=1}^n\frac{1}{1-pz} \\ \times
\sum_{q=0}^{2n-4} {2n-4\choose q} z^q
\sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-j} (j+1)^q; z=0\right)$$
which is
$$A_n =
- \frac{1}{2} \mathrm{Res}
\left(\frac{1}{z^n}
\prod_{p=1}^n\frac{1}{1-pz} \times
\sum_{q=0}^{2n-4} {2n-4\choose q} {q+1\brace n} z^q; z=0\right).$$
which simplifies to (powers $z^q$ with $q\ge n$ do not contribute to the residue)
$$A_n =
- \frac{1}{2} \mathrm{Res}
\left(\frac{1}{z^n}
\prod_{p=1}^n\frac{1}{1-pz} \times
\sum_{q=0}^{n-1} {2n-4\choose q} {q+1\brace n} z^q; z=0\right)$$
which is (the Stirling number vanishes when $q+1 < n$)
$$A_n =
- \frac{1}{2} \mathrm{Res}
\left(\frac{1}{z^n}
\prod_{p=1}^n\frac{1}{1-pz} \times
{2n-4\choose n-1} {n\brace n}z^{n-1};z=0\right)
= -\frac{1}{2} {2n-4\choose n-1}.$$
For $B_n$ we obtain
$$B_n =
\mathrm{Res}
\left(\frac{1}{z^n}
\prod_{p=1}^n\frac{1}{1-pz} \times
\sum_{q=0}^{2n-4} {2n-4\choose q} {q+2\brace n} z^{q+1}; z=0\right)$$
which simplifies to
$$B_n =
\mathrm{Res}
\left(\frac{1}{z^n}
\prod_{p=1}^n\frac{1}{1-pz} \times
\sum_{q=0}^{n-2} {2n-4\choose q} {q+2\brace n} z^{q+1}; z=0\right)$$
which is
$$B_n =
\mathrm{Res}
\left(\frac{1}{z^n}
\prod_{p=1}^n\frac{1}{1-pz} \times
{2n-4\choose n-2} {n\brace n}z^{n-1};z=0\right)
= {2n-4\choose n-2}.$$
For $C_n$ we obtain
$$C_n =
- \frac{1}{2} \mathrm{Res}
\left(\frac{1}{z^n}
\prod_{p=1}^n\frac{1}{1-pz} \times
\sum_{q=0}^{2n-4} {2n-4\choose q} {q+3\brace n} z^{q+2}; z=0\right)$$
which simplifies to
$$C_n =
- \frac{1}{2} \mathrm{Res}
\left(\frac{1}{z^n}
\prod_{p=1}^n\frac{1}{1-pz} \times
\sum_{q=0}^{n-3} {2n-4\choose q} {q+3\brace n} z^{q+2}; z=0\right)$$
which is
$$C_n =
- \frac{1}{2} \mathrm{Res}
\left(\frac{1}{z^n}
\prod_{p=1}^n\frac{1}{1-pz} \times
{2n-4\choose n-3} {n\brace n}z^{n-1};z=0\right)
= - \frac{1}{2} {2n-4\choose n-3}.$$
Finally collecting the contributions from $A_n, B_n$ and $C_n$ we
obtain
$$-\frac{1}{2} {2n-4\choose n-1}
+ {2n-4\choose n-2}
- \frac{1}{2} {2n-4\choose n-3}
\\ =
\left(-\frac{1}{2} \frac{n-2}{n-1}
+ 1 - \frac{1}{2} \frac{n-2}{n-1} \right)
{2n-4\choose n-2}
= \frac{1}{n-1} {2n-4\choose n-2} = C_{n-2}.$$
Addendum. Here we have used the basic Stirling number identity
$$\sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-j} (j+1)^q = - (n-1)! {q+1\brace n}.$$
To prove this consider the generating function
$$- \sum_{q\ge 0} \frac{z^q}{q!} \frac{1}{(n-1)!}
\sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-j} (j+1)^q
\\= \frac{1}{(n-1)!} \sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-1-j} \sum_{q\ge 0} \frac{z^q}{q!} (j+1)^q
\\= \frac{1}{(n-1)!} \sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-1-j} \exp((j+1)z)
\\= \frac{\exp(z)}{(n-1)!} \sum_{j=0}^{n-1} {n-1\choose j}
(-1)^{n-1-j} \exp(jz)
\\= \frac{\exp(z)}{(n-1)!} (\exp(z)-1)^{n-1}.$$
Now observe the exponential generating function
$$\sum_{q\ge 0} {q+1\brace n} \frac{z^q}{q!} =
\left(\sum_{q\ge 0} {q\brace n} \frac{z^q}{q!}\right)'$$
which is
$$ \left(\frac{1}{n!} (\exp(z)-1)^n\right)'
= \frac{1}{n!} \times n \times (\exp(z)-1)^{n-1} \exp(z)
\\ = \frac{1}{(n-1)!} (\exp(z)-1)^{n-1} \exp(z).$$
The two generating functions are the same, fin.
Here we have used the bivariate generating function
$$G(z, u) = \exp(u(\exp(z)-1))$$
of the Stirling numbers of the second kind which follows from the
combinatorial class specification
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}(\mathcal{U}\times\textsc{SET}_{\ge 1}(\mathcal{Z})).$$
Here is a possible line of attack that will give the desired result if we can prove the following (true) lemma
Lemma: For all $n>1$
$$\sum_{k=1}^{n-1} \frac{1}{k^{3/2}(n-k)^{3/2}} \leq \frac{2}{n^{3/2}}$$
Assume $C_k \leq \frac{4^k}{k^{3/2}}$ for $k=1,2,\ldots, n$ then
$$C_{n+1} \leq 4^{n}\left(\frac{2}{n^{3/2}} + \sum_{k=1}^{n-1} \frac{1}{k^{3/2}(n-k)^{3/2}}\right) \leq \frac{4^{n+1}}{(n+1)^{3/2}}$$
by the above lemma and the assumed bound on $C_k$ follows by induction. This result gives
$$\left|\sum_{k=0}^n C_k x^k\right| \leq 1 + \sum_{k=1}^n \frac{(4x)^k}{k^{3/2}}$$
and it follows that the series converges for $|x| < \frac{1}{4}$.
Best Answer
Here's an idea that seems promising but didn't quite work out the way I expected; maybe there will be something of use in it? The basic strategy is that, since you know what you want when $r = 0$, it would be enough to show that $f(m + 1, r + 1) - f(m, r)$ has the correct value (and I choose this particular change in inputs to keep the difference in the right sides $[c^{2m + 1 - 2r}]_r$ tame). So here goes (apologies in advance for the horrendously large formulas):
Let $\displaystyle f(m, r) := \sum_{j=r}^{2m}\left(\binom{2m}{j}-\binom{2m}{2r-j-1}\right)x^j c(x)^{2j+1-2r}$. Since $x c^2 = c - 1$, we can rewrite this as $$f(m, r) = \sum_{j=r}^{2m}\left(\binom{2m}{j}-\binom{2m}{2r-j-1}\right)(c - 1)^j c^{1-2r}.$$ Now let's consider $f(m + 1, r + 1)$; by definition, we have \begin{align*} c^{2r + 1} f(m + 1, r + 1) & = \sum_{j=r + 1}^{2m + 2}\left(\binom{2m + 2}{j}-\binom{2m + 2}{2r-j + 1}\right)(c - 1)^j \\ & = \sum_{j=r + 1}^{2m + 2}\left(\binom{2m}{j} + 2\binom{2m}{j - 1} + \binom{2m}{j - 2} -\binom{2m}{2r-j + 1} - 2 \binom{2m}{2r - j} - \binom{2m}{2r - j - 1} \right)(c - 1)^j \\ & = \sum_{j=r + 1}^{2m + 2}\left(\binom{2m}{j - 2} -\binom{2m}{2r-j + 1} \right)(c - 1)^j +2\sum_{j=r + 1}^{2m + 2}\left(\binom{2m}{j - 1} - \binom{2m}{2r-j}\right)(c - 1)^j +\sum_{j=r + 1}^{2m + 2}\left(\binom{2m}{j} -\binom{2m}{2r-j - 1}\right)(c - 1)^j \\ & = \sum_{j=r - 1}^{2m}\left(\binom{2m}{j} -\binom{2m}{2r-j - 1} \right)(c - 1)^{j + 2} +2\sum_{j=r}^{2m + 1}\left(\binom{2m}{j} - \binom{2m}{2r-j - 1}\right)(c - 1)^{j + 1} +\sum_{j=r + 1}^{2m + 2}\left(\binom{2m}{j} -\binom{2m}{2r-j - 1}\right)(c - 1)^j.\end{align*} In the last expression, the first sum is $c^{2r - 1}(c - 1)^2 \left(f(m, r) + \left(\binom{2m}{r - 1} - \binom{2m}{r}\right)(c - 1)^{r - 1}\right)$, the second sum is $2c^{2r - 1}(c - 1)f(m, r)$ and the third sum is $c^{2r - 1} \left( f(m, r) - \left(\binom{2m}{r} - \binom{2m}{r - 1}\right)(c - 1)^r\right)$. Therefore \begin{align*} c^{2r + 1} f(m + 1, r + 1)& = c^{2r - 1}\left(\left((c - 1)^2 + 2(c - 1) + 1\right)f(m, r) + \left(\binom{2m}{r - 1} - \binom{2m}{r}\right)(c - 1)^{r + 1} - \left(\binom{2m}{r} - \binom{2m}{r - 1}\right)(c - 1)^r\right) \\ & = c^{2r + 1} f(m, r)+ c^{2r - 1}(c - 1)^r\left(\binom{2m}{r - 1} - \binom{2m}{r}\right)((c - 1) + 1)\end{align*} and therefore $$f(m + 1, r + 1) - f(m, r) = c^{-1}(c - 1)^r\left(\binom{2m}{r - 1} - \binom{2m}{r}\right).$$ Now I was hoping to get something that was obviously just a single monomial in $x$, which obviously this is not. But maybe it's just a small error somewhere?