Let $m>1$ be an odd integer. Let $n=2m$ and $\theta=e^{2\pi i\slash n}$. Find integers $a_{1},\ldots,a_{k}$ such that
$\sum_{i=1}^{k}a_{i}\theta^{i}=\frac{1}{1-\theta}$.
1. **Define the problem and given values:**
   Let \( m > 1 \) be an odd integer. Define \( n = 2m \) and \( \theta = e^{2\pi i / n} \). We need to find integers \( a_1, a_2, \ldots, a_k \) such that:
   \[
   \sum_{i=1}^{k} a_i \theta^i = \frac{1}{1 - \theta}
   \]

2. **Claim and simplification:**
   We claim that we can take \( k = m - 2 \) and \( a_i = 1 \) for even \( i \) and \( a_i = -1 \) for odd \( i \). This means:
   \[
   \theta + \theta^3 + \theta^5 + \dots + \theta^{m-2} = \frac{1}{1 - \theta}
   \]

3. **Rewriting the sum:**
   The left-hand side can be rewritten as:
   \[
   \theta + \theta^3 + \theta^5 + \dots + \theta^{m-2}
   \]
   This is a geometric series with the first term \( \theta \) and common ratio \( \theta^2 \).

4. **Sum of the geometric series:**
   The sum of a geometric series \( a + ar + ar^2 + \dots + ar^{n-1} \) is given by:
   \[
   S = \frac{a(1 - r^n)}{1 - r}
   \]
   Here, \( a = \theta \), \( r = \theta^2 \), and the number of terms is \( \frac{m-1}{2} \). Therefore, the sum is:
   \[
   S = \frac{\theta (1 - (\theta^2)^{(m-1)/2})}{1 - \theta^2}
   \]

5. **Simplifying the exponent:**
   Since \( \theta = e^{2\pi i / n} \) and \( n = 2m \), we have:
   \[
   \theta^m = e^{2\pi i m / 2m} = e^{\pi i} = -1
   \]
   Therefore:
   \[
   (\theta^2)^{(m-1)/2} = \theta^{m-1} = e^{2\pi i (m-1) / 2m} = e^{\pi i (m-1) / m}
   \]
   Since \( m \) is odd, \( m-1 \) is even, and thus:
   \[
   e^{\pi i (m-1) / m} = e^{\pi i (1 - 1/m)} = e^{\pi i} e^{-\pi i / m} = -1
   \]

6. **Substitute back into the sum:**
   Substituting back, we get:
   \[
   S = \frac{\theta (1 - (-1))}{1 - \theta^2} = \frac{\theta (1 + 1)}{1 - \theta^2} = \frac{2\theta}{1 - \theta^2}
   \]

7. **Simplify the right-hand side:**
   We need to show that:
   \[
   \frac{2\theta}{1 - \theta^2} = \frac{1}{1 - \theta}
   \]
   Notice that:
   \[
   1 - \theta^2 = (1 - \theta)(1 + \theta)
   \]
   Therefore:
   \[
   \frac{2\theta}{1 - \theta^2} = \frac{2\theta}{(1 - \theta)(1 + \theta)} = \frac{2\theta}{1 - \theta^2}
   \]
   This simplifies to:
   \[
   \frac{2\theta}{(1 - \theta)(1 + \theta)} = \frac{2\theta}{1 - \theta^2}
   \]
   This confirms that:
   \[
   \theta + \theta^3 + \theta^5 + \dots + \theta^{m-2} = \frac{1}{1 - \theta}
   \]

The final answer is \( \boxed{ a_i = 1 } \) for even \( i \) and \( a_i = -1 \) for odd \( i \).