Let $\binom{\vec{a}}{\vec{b}}$, with $n$-dimensional lists of nonnegative integers $\vec{a}$ and $\vec{b}$ denotes the product of binomial coefficients: $\binom{\vec{a}}{\vec{b}}=\prod^n_{j=1}\binom{a_j}{b_j}$, where the factors with $a_j=0$ are removed (assumed $\binom{0}{b_j}=1$). Further, $\vec{c}(n,m)$, stands for a sequence of $n$ nonnegative integers, $\vec{c}(n,m)=(c_1,\dots,c_n)$, such that $\sum_{j=1}^n jc_j=m$ (if $n>m$ obviously $c_j=0$ for $j>m$). For even $N$ I would like to calculate:
$$X=\sum_{\vec{c}(N,N/2)}\binom{\vec{c}(N,N)}{\vec{c}(N,N/2)}$$ (fixed $\vec{c}(N,N)$, while sum runs over all possible $\vec{c}(N,N/2)$).
$$Y=\sum_{\vec{c}(N,N/2)}\binom{\vec{c}(N,N)}{\vec{c}(N,N/2)}-\sum_{\vec{c}(N,N/2+1)}\binom{\vec{c}(N,N)}{\vec{c}(N,N/2+1)}.$$
The problem appears in the context of the representations of symmetric group. I think that $Y$ is the character of the irreducible representation $\lambda=(N/2,N/2)$ (Young scheme) of $S_N$ for the permutation with cyclic structure $\vec{c}(N,N)$, i.e. with $c_j$ cycles of length $j$. If the closed form for this is found, I will be happy to get a reference.
If the task means that $\vec{c}(n)$ is an $n$-tuple of positive integers such that $\sum_{j=1}^n jc_n=n$ (that's how it is described), then $c_n(n)(n+1)/2=n$, or equivalently, $c_n=\frac{2}{n+1}$. But then the task is ill-defined for all $n>1$.
If it is meant that $\sum_j jc_j=n$ and all $c_j$ are positive (as stated), then the task is still ill-defined since this is impossible to satisfy, except for $n=1$.
If it is meant that $\sum_j jc_j=n$ and all $c_j$ are non-negative, then $(c_1,\ldots,c_n)$ forms an integer partition of $n$. In this case, $\sum_{\vec{c}(N/2)}$ sums over all integer partitions of $N/2$. In this case, I think it will be difficult to get closed-form solutions except for some very simple $\vec{c}(N)$.