Quite interesting! Claim 1: $\sum_{i=1}^n a_i^{2k+1} = 0$ for all $k \in \mathbb{N}$ Let $M$ be an $n \times n$ matrix with rows counted from $0$ to $n-1$ and row $i$ is $(a_1^{2i}, a_2^{2i}, \dots , a_n^{2i})$. We have that for the vector $v = (a_1, a_2, \dots, a_n)$ \[ M \cdot v = 0 \]Since $v$ is not the 0-vector, the rows in $M$ must be linearly dependent. We now proceed by induction: assume the claim holds for all $0 \leq k < r$. We want to prove that $\sum_{i=1}^n a_i^{2r+1} = 0$ Becaue the rows of $M$ are linearly dependent there exists some $0<m\leq n$ such that row $m$ is a linear combination of rows $0,1,\dots ,m-1$. Now consider: \[ \sum_{i=1}^{n} a_i^{2r+1} = (a_1^{2(r-m)+1}, a_2^{2(r-m)+1}, \, \, \dots \, \, , a_n^{2(r-m)+1}) \, \, \cdot \,\, (a_1^{2m+1}, a_2^{2m+1}, \dots ,a_n^{2m+1})\]However, row $m$ can be split into its linear combinations. \[ a_i^{2m+1} = \sum_{j=1}^{m-1} C_ja_i^{2j+1} \]For some constants $C_1, C_2, \dots C_{m-1}$ For any row $k < m$ we have: \[ (a_1^{2(r-m)+1}, a_2^{2(r-m)+1}, \dots, a_n^{2(r-m)+1}) \, \, \cdot \, \, C_k(a_1^{2k+1}, a_2^{2k+1}, \dots , a_n^{2(k+1)}) = C_k\sum_{i=1}^n a_i^{2(r-k)+1} = 0 \]Because of our induction case. Summing up the different components of row $m$ we get that the product must be 0, since all components are also 0. Thus our claim holds for $r$ and thus for any $r \in \mathbb{N}$ Now assume WLOG that all of the $a_i$ are non-zero. Let the sequence $b_i$ denote the absolute value of all the negative terms of $a_i$ and $c_i$ denote all the positive terms of $a_i$ such that: \[ 0 < b_1 \leq b_2 \leq \dots \leq b_x \]\[ 0 < c_1 \leq c_2 \leq \dots \leq c_y \]for $x+y=n$ (So we have $b_x = -a_1$, $b_{x-1} = -a_2$ and so on as long as the $a_i$ are negative) Thus we must have for all positive integers $k$: \[ \sum_{i=1}^x b_i^{2k+1} = \sum_{j=1}^y c_i^{2k+1} \] Claim 2: Both sequences are identical Let $f(k) = \sum_{i=1}^x b_i^{2k+1}$ and $g(k) = \sum_{i=1}^y c_i^{2k+1}$ Clearly both $f$ and $g$ are strictly increasing. Assume for the sake of contradiction that $b_x > c_y$. It follows that $b_x^{2K+1} > g(K)$ for all $K$ greater than some arbitrary large bound. However this is impossible, since $b_x^{2K+1} < f(K) = g(K)$ for all $K$. Arguing similarly for $c_y$ we find that $c_x = b_y$. This allows us to "delete" these terms and continue the argument for all smaller elements of the sequences. In the end the claim is shown. Now because of how we've defined $b_i$ and $c_i$ we must have that $b_i$ represents the negative value of $c_i$ in the original $a_i$ sequence. Thus the claim of the problem obviously follows and we are done.