Find an expression for $f_j(x) = {2N \choose N}^{-1}\sum_i {2j \choose i}{2N-2j \choose N-i} x^i$. Background: I am trying to find the values of the following two expressions
$$
a_j = \sum_{i=0}^{2j} i\frac{{2j \choose i}{2N-2j \choose N-i}}{2N\choose N}
$$
and 
$$
b_j = \sum_{i=0}^{2j} i^2\frac{{2j \choose i}{2N-2j \choose N-i}}{2N\choose N}.
$$
To do this, I defined the function
$$
f_j(x) = {2N \choose N}^{-1}\sum_{i=0}^{2j} {2j \choose i}{2N-2j \choose N-i} x^i.
$$
If I can sum this expression, the values of $a_j$ and $b_j$ can be computed by differentiating $f_j$ with respect to $x$. But I have no idea where to start to find $f_j$. 
Edit: I'm guessing $a_j = j$, but I doubt it will help finding $f_j$.
Edit: Also it looks like that $b_j-j^2 = \frac{j(N-j)}{2N-1}$
 A: Only a suggestion.
The summation bounds are actually superfluos, because they are implicit in the Binomials. So we can write:
$$
\eqalign{
  & \sum\limits_{0\, \le \,i\, \le \,2j\,} {\left( \matrix{
  2j \cr 
  i \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - i \cr}  \right)x^{\,i} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,\min \left( {N,\;2j} \right)} \right)\,} {\left( \matrix{
  2j \cr 
  i \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - i \cr}  \right)x^{\,i} }  \cr} 
$$
We must here pay attention to the fact that the above is not the Vandermonde identity as suggested in a comment.
 And not because restricted by the sum limits, while instead because the Vandermonde identity reads
$$
\eqalign{
  & \left( {1 + x} \right)^{\,2N}  = \left( {1 + x} \right)^{\,2N - 2j} \left( {1 + x} \right)^{\,2j}  =   \cr 
  &  = \sum\limits_i {\left( \matrix{
  2j \cr 
  i \cr}  \right)x^{\,i} } \sum\limits_k {\left( \matrix{
  2N - 2j \cr 
  k \cr}  \right)x^{\,k} }  =   \cr 
  &  = \sum\limits_s {\left( {\sum\limits_i {\left( \matrix{
  2j \cr 
  i \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  s - i \cr}  \right)} } \right)x^{\,s} }  =   \cr 
  &  = \sum\limits_s {\left( \matrix{
  2N \cr 
  s \cr}  \right)x^{\,s} }  \cr} 
$$
The sum is not easy to manage and requires some more thinking.
However the sum in $a_j$ can be solved quite easily to give
$$
\eqalign{
  & a(j) = \sum\limits_{\left( {0\, \le } \right)\,i\,\,\left( { \le \,\min \left( {N,\;2j} \right)} \right)\,} {i\left( \matrix{
  2j \cr 
  i \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - i \cr}  \right)}  =   \cr 
  &  = 2j\sum\limits_{\left( {0\, \le } \right)\,i\,\,\left( { \le \,\min \left( {N,\;2j} \right)} \right)\,} {\left( \matrix{
  2j - 1 \cr 
  i - 1 \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - i \cr}  \right)}  =   \cr 
  &  = 2j\sum\limits_{\left( {0\, \le } \right)\,i\,\,\left( { \le \,\min \left( {N,\;2j} \right)} \right)\,} {\left( \matrix{
  2j - 1 \cr 
  i - 1 \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - 1 - \left( {i - 1} \right) \cr}  \right)}  =   \cr 
  &  = 2j\sum\limits_{\left( {0\, \le } \right)\,k\,\,\left( { \le \,\min \left( {N - 1,\;2j - 1} \right)} \right)\,} {\left( \matrix{
  2j - 1 \cr 
  k \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - 1 - k \cr}  \right)}  =   \cr 
  &  = 2j\left( \matrix{
  2N - 1 \cr 
  N - 1 \cr}  \right) \cr} 
$$
And for $b_j$
$$
\eqalign{
  & b(j) = \sum\limits_{\left( {0\, \le } \right)\,i\,\,\left( { \le \,\min \left( {N,\;2j} \right)} \right)\,} {i^{\,2} \left( \matrix{
  2j \cr 
  i \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - i \cr}  \right)}  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,i\,\,\left( { \le \,\min \left( {N,\;2j} \right)} \right)\,} {\left( {i\left( {i - 1} \right) + i} \right)\left( \matrix{
  2j \cr 
  i \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - i \cr}  \right)}  \cr} 
$$
i.e.
$$
\eqalign{
  & b(j) - a(j) = \sum\limits_{\left( {0\, \le } \right)\,i\,\,\left( { \le \,\min \left( {N,\;2j} \right)} \right)\,} {i\left( {i - 1} \right)\left( \matrix{
  2j \cr 
  i \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - i \cr}  \right)}  =   \cr 
  &  = 2j\left( {2j - 1} \right)\sum\limits_{\left( {0\, \le } \right)\,i\,\,\left( { \le \,\min \left( {N,\;2j} \right)} \right)\,} {\left( \matrix{
  2j - 2 \cr 
  i - 2 \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - i \cr}  \right)}  =   \cr 
  &  = 2j\left( {2j - 1} \right)\sum\limits_{\left( {0\, \le } \right)\,i\,\,\left( { \le \,\min \left( {N,\;2j} \right)} \right)\,} {\left( \matrix{
  2j - 2 \cr 
  i - 2 \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - 2 - \left( {i - 2} \right) \cr}  \right)}  =   \cr 
  &  = 2j\left( {2j - 1} \right)\sum\limits_{\left( {0\, \le } \right)\,k\,\,\left( { \le \,\min \left( {N - 2,\;2j - 2} \right)} \right)\,} {\left( \matrix{
  2j - 2 \cr 
  k \cr}  \right)\left( \matrix{
  2N - 2j \cr 
  N - 2 - k \cr}  \right)}  =   \cr 
  &  = 2j\left( {2j - 1} \right)\left( \matrix{
  2N - 2 \cr 
  N - 2 \cr}  \right) \cr} 
$$
Inserting the normalization constant $\binom{2N}{N}$ we get
$$
{{a(j)} \over {\left( \matrix{
  2N \cr 
  N \cr}  \right)}} = 2j{{\left( \matrix{
  2N - 1 \cr 
  N - 1 \cr}  \right)} \over {\left( \matrix{
  2N \cr 
  N \cr}  \right)}} = 2j{{N!} \over {\left( {N - 1} \right)!\left( {2N} \right)}} = j\quad \left| {\,1 \le N} \right.
$$
and
$$
\eqalign{
  & {{b(j)} \over {\left( \matrix{
  2N \cr 
  N \cr}  \right)}} = {{a(j)} \over {\left( \matrix{
  2N \cr 
  N \cr}  \right)}} + 2j\left( {2j - 1} \right){{\left( \matrix{
  2N - 2 \cr 
  N - 2 \cr}  \right)} \over {\left( \matrix{
  2N \cr 
  N \cr}  \right)}} =   \cr 
  &  = j + 2j\left( {2j - 1} \right){{N\left( {N - 1} \right)} \over {\left( {2N} \right)\left( {2N - 1} \right)}} =   \cr 
  &  = j + j\left( {2j - 1} \right){{\left( {N - 1} \right)} \over {\left( {2N - 1} \right)}} =   \cr 
  &  = j^{\,2}  + {{j\left( {N - j} \right)} \over {\left( {2N - 1} \right)}}\quad \left| {\;1 \le N} \right. \cr} 
$$
which confirms your hypotheses.
