Let $\displaystyle P(z) = \sum_{k=0}^n (-1)^ke_k z^{n-k}$ be the $n^{th}$ degree polynomial with roots $z_k \mid k=1,\cdots,n$, where by Vieta's formulas $e_k$ are the elementary symmetric polynomials.
Let $\displaystyle p_i=\sum_{k=1}^n z_k^i\,$, where it is given that $p_l=0$ for all odd $l$.
From Newton's identities $\displaystyle k e_k = \sum_{i=1}^k (-1)^{i-1}e_{k-i}p_i$ it follows (by induction, for example) that $e_l=0$ for all odd $l$. Therefore, the polynomial $P(z)$ has every other coefficient $0$, so it contains either only even powers of $z$, or only odd powers of $z$, depending on the parity of $n$. In the first case $P(z)$ is an even function, in the second case an odd one. In both cases $P(z)=0 \iff P(-z)=0$ so the roots of $P(z)$ can be grouped in pairs of mutually opposites.
We prove instead the following Lemma. Using this Lemma, your claim follows by an immediate (strong) induction.
Lemma Let $a_1,.., a_k$ be complex numbers, not all of them zero, and $z_1,..,z_k$ non-zero, pairwise distinct complex numbers. If $$a_1z_1^l+...+a_kz_k^l=0$$ for all odd integers $l$, then, there exists some $i\neq j$ such that $$z_i+z_j=0$$
Proof:
Consider the determinant $$\Delta=\begin{vmatrix} z_1 & z_2 & z_3 &...&z_k \\ z_1^3& z_2^3 & z_3^3 &...&z_k^3 \\ z_1^5 & z_2^5 & z_3^5 &...&z_k^5 \\ ...&...&...&....&... \\ z_1^{2k-1} & z_2^{2k-1} & z_3^{2k-1} &...&z_k^{2k-1} \\ \end{vmatrix}$$
First, since $a_1 \mbox{col 1}+...+a_k \mbox{col k}=0$ we get $\Delta=0$. Next, using the Vandermonde formula, we get $$0=\Delta=z_1z_2...z_k\begin{vmatrix} 1 & 1 & 1 &...&1 \\ z_1^2& z_2^2 & z_3^2 &...&z_k^2 \\ z_1^4 & z_2^4 & z_3^4 &...&z_k^4 \\ ...&...&...&....&... \\ z_1^{2k-2} & z_2^{2k-2} & z_3^{2k-2} &...&z_k^{2k-2} \\ \end{vmatrix}\\=z_1z_2..z_k \prod_{1 \leq i <j \leq k} (z_j^2-z_i^2)$$
It follows from the hypothesis that there exists some $i <j$ such that $z_j^2-z_i^2=0$ and hence $z_i+z_j=0$.
