The vanishing of sum of coefficients: symmetric polynomials Denote $\pmb{X}_n=(x_1,x_2,\dots,x_n)$. Consider the symmetric polynomial
$$f_n(\pmb X_n)=\prod_{1\leq i<j\leq n}(x_i+x_j).$$
Expand these in terms of elementary symmetric polynomials, say
$$f_n(\pmb{X}_n)=\sum_{\mu}c_{\mu,n}\cdot e_{\mu}(\pmb{X}_n).$$
For example,
\begin{align*} f_3&=-e_{(3)}+e_{(2,1)} \\
f_5&=-e_{(5,5)}+2e_{(5,4,1)}+e_{(5,3,2)}-e_{(5,2,2,1)}-e_{(4,4,1,1)}-e_{(4,3,3)}+e_{(4,3,2,1)}.
\end{align*}

QUESTION 1. Is it true that, for integers $n \geq 1$, we have
  $$\sum_{\mu}c_{\mu,2n+1}=0?$$

POSTSCRIPT. Fedor's reply (to Question 1) shown below suggests to me to ask:

QUESTION 2. Is it true that, for integers $n \geq 1$, we have
  $$\sum_{\mu}c_{\mu,2n}=(-1)^{\binom{n}2}?$$

 A: Here is another method using more of the theory of symmetric
functions. By Enumerative Combinatorics, vol. 2, Exercise 7.30, we have
$f_n(\boldsymbol{X}_n)= s_{(n-1,n-2,\dots,1)}(\boldsymbol{X}_n)$ (a
Schur function). By the dual Jacobi-Trudi identity,
  $$ s_{(n-1,n-2,\dots,1)} = \det[e_{n-2i+j}]_{i,j=1}^{n-1},\ \ (*)  $$
where $e_0=1$ and $e_k=0$ for $k<0$. Since $e_k(\boldsymbol{X}_n)=0$
for $k>n$, it follows that $\sum_\mu c_{\mu,n}$ is obtained by
substituting $e_1=e_2=\cdots=e_n=1$ and $e_{n+1}=e_{n+2}=\cdots=0$
into the right-hand side of (*). When $n$ is odd, the two middle rows
of the determinant are equal (in fact, they are all 1's), so the
determinant is 0. If $n=2m$ then subtract row $m-1$ from row $m$, then
row $m-2$ from row $m-1$, up to row 1 from row 2. Also subtract row
$m+2$ from row $m+1$,  row $m+3$ from $m+2$, etc. The resulting matrix $A$
can be transformed into a triangular matrix $B$ with 1's on the diagonal
by row and column permutations. The permutation indexing the 1's in $A$ that become the diagonal elements of $B$ is
$1,3,5,\dots,n-1,2,4,6,\dots,n-2$, which has ${m\choose 2}$
inversions, and the proof follows.
A: Choose $n$ numbers $x_1,\dots,x_n$ for which all elementary symmetric polynomials are equal to 1 and substitute them to our $f_n$. We should get zero value for odd $n$. Well, what are these numbers? The roots of $x^{n}-x^{n-1}+x^{n-2}-\ldots-1=(x^{n+1}-1)/(x+1)$. This polynomial indeed has two roots with sum equal to 0 when $n$ is odd.
If $n=2k$ is even, we substitute the roots $w_1,\dots,w_n$ of the polynomial $f(x)=x^{2k}-x^{2k-1}+\ldots+1=(x^{2k+1}+1)/(x+1)=(x-w_1)\dots (x-w_n)$. Then your claim reads as $$A:=\prod_{1\leqslant i<j\leqslant n} (w_i+w_j)=(-1)^{\binom{k}2}.$$ This is done by the standard trick (and is well known itself). At first, 
$$
|A|^2=\prod_{i=1}^n \prod_{j\ne i,1\leqslant j\leqslant n}|w_i+w_j|=2^{-n}\prod_{i=1}^n \prod_{j=1}^n|(-w_i)-w_j|=2^{-n}\prod_{i=1}^n |f(w_i)|=\\=2^{-n}\prod_{i=1}^n\left|\frac{(-w_i)^{2k+1}+1}{-w_i+1}\right|=1,
$$
since $1+(-w_i)^{2k+1}=2$ for all $i=1,2,\dots,n$ and $\prod_{i=1}^n (1-w_i)=f(1)=1$.
At second, we need to find the argument of the complex number $A$. This may be done for example as follows: all pairs $w_i+w_j$ for which $w_i$ and $w_j$ are not complex conjugate are partitioned onto complex conjugate pairs. In each pair the product is positive reals. If $w_i$ and $w_j$ are complex conjugate, the sum $w_i+w_j$ is a real number whose sign is the sign of the real part of $w_i$. Therefore $A$ is the real number whose sign equals $(1)^{m/2}$, where $m$ is the number of $w$'s in the left half-plane. It is easy to see that $m/2=[k/2]$ and that $(-1)^{[k/2]}=(-1)^{k(k-1)/2}$.
A: The polynomial in question is an instance of the Boolean product polynomials,
which might give some extra insight. For example, I believe Lascoux have studied the Schur expansion of that exact expression.
