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 $b_j-j^2 = \frac{j(N-j)}{2N-1}$
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.