On Vanishing Riemann Sums and Odd Functions 
Let $ f: [-1,1] \to \mathbb{R} $ be a continuous function. Suppose that the $ n $-th midpoint Riemann sum of $ f $ vanishes for all $ n \in \mathbb{N} $. In other words,
  $$
\forall n \in \mathbb{N}: \quad \mathcal{R}^{f}_{n} := \sum_{k=1}^{n} f \left( -1 + \frac{2k - 1}{n} \right) \cdot \frac{2}{n} = 0.
$$
  Question: Is it necessarily true that $ f $ is an odd function?

It is easy to verify that if $ f $ is an odd continuous function, then $ \mathcal{R}^{f}_{n} = 0 $ for all $ n \in \mathbb{N} $. However, is the converse true?
This is part of an original research problem, so unfortunately, there is no other source except myself. With someone else, I managed to obtain the following partial result.

Theorem If $ f $ is a polynomial function and $ \mathcal{R}^{f}_{n} = 0 $ for all $ n \in \mathbb{N} $, then $ f $ has only odd powers, which immediately implies that $ f $ is an odd function.

The proof relies on properties of Bernoulli polynomials and Vandermonde matrices.
For the general case, I was thinking that Fourier-analytic tools might help, such as Poisson summation. A Fourier-analytic approach seems promising, but it has limitations and might not be able to fully resolve the question.
Would anyone care to offer some insight into the problem? Thanks!
 A: Take the function $f(x)=\sum_{j\geq1} \alpha_j \cos(\pi j x)$. Then its $n$-th midpoint Riemann sum is
$$\begin{align}
 0 = R_n f &= 
\sum_{j\geq1}\alpha_j \sum_{1\leq k\leq n}\frac{2}{n} \cos\left( \pi j\left( -1 + \frac{2k-1}{n}\right)\right)
\\&= \frac{2}{n}\sum_{j\geq1}\alpha_j \sum_{1\leq k\leq n} (-1)^j\cos\left(\pi j(2k-1)/n\right)
\\&= \frac{2}{n} \sum_{j\geq1} \alpha_j \frac{\sin \pi j}{\sin (\pi j/n)}
\end{align}$$
where (by Mathematica) 
$$ \sum_{1\leq k\leq n}\cos\frac{\pi j(2k-1)}{n} = \frac{\cos \pi j\sin \pi j}{\sin (\pi j/n)}$$
and when I write $\sin \pi j/\sin(\pi j/n)$ I mean the limit as $j$ approaches its integer value (so no division by zero).
Now, when $n=1$, the condition is
$$ 0 = \sum_{j\geq1} \alpha_j $$
and when $n>1$, the condition is
$$ 0 = \sum_{j\geq1} \alpha_j(-1)^{(j/n)}[n\backslash j]. $$
The condition $R_n f=0$ is only nontrivial when there are $j$ such that $n\backslash j$ and $\alpha_j\neq0$. So suppose that $\alpha_j\neq0$ only when $j$ is a power of 2, so that the function is
$$ f(x) = \sum_{k\geq0} \beta_k \cos(\pi 2^k x). $$
Then the only $n$ that impose any conditions on $\alpha_k$ are the powers of 2.
If $n=2^m$, $m>0$, then the condition is
$$ \beta_m - \beta_{m+1}-\beta_{m+2}-\cdots = 0, $$
and for $n=1$ the condition is
$$ \sum_{k\geq0} \beta_k = 0. $$
Pick $\beta_0 = -1, \beta_k = 2^{-k}$ ($k\geq1$). 
The condition for each $n=2^m$ and also $n=1$ will be satisfied, the function
$$ f(x) = -\cos\pi x+\sum_{k\geq1} 2^{-k} \cos(\pi 2^k x) $$
is clearly even and nonzero, and $R_nf=0$ for every $n$.
If the Fourier series is finite, the function must then be zero.
