Find all natural numbers $n$ for which there is a permutation $\sigma$ of $\{1,2,\ldots, n\}$ that satisfies:
\[
\sum_{i=1}^n \sigma(i)(-2)^{i-1}=0
\]
To solve the problem, we need to find all natural numbers \( n \) for which there exists a permutation \( \sigma \) of \(\{1,2,\ldots, n\}\) that satisfies:
\[
\sum_{i=1}^n \sigma(i)(-2)^{i-1}=0
\]

### Step 1: Analyze the problem modulo 3
First, we analyze the given equation modulo 3. Notice that:
\[
(-2)^{i-1} \equiv (-1)^{i-1} \pmod{3}
\]
This is because \((-2) \equiv 1 \pmod{3}\).

### Step 2: Simplify the sum modulo 3
Thus, the equation becomes:
\[
\sum_{i=1}^n \sigma(i)(-1)^{i-1} \equiv 0 \pmod{3}
\]

### Step 3: Sum of the first \( n \) natural numbers
The sum of the first \( n \) natural numbers is:
\[
\sum_{i=1}^n i = \frac{n(n+1)}{2}
\]

### Step 4: Check \( n \equiv 1 \pmod{3} \)
Assume \( n \equiv 1 \pmod{3} \). Then:
\[
\frac{n(n+1)}{2} \equiv \frac{1 \cdot 2}{2} \equiv 1 \pmod{3}
\]
This implies:
\[
\sum_{i=1}^n \sigma(i)(-1)^{i-1} \equiv 1 \pmod{3}
\]
which contradicts the requirement that the sum is 0 modulo 3. Therefore, \( n \equiv 1 \pmod{3} \) does not work.

### Step 5: Check \( n \equiv 0 \pmod{3} \) and \( n \equiv 2 \pmod{3} \)
We need to show that \( n \equiv 0 \pmod{3} \) and \( n \equiv 2 \pmod{3} \) work. We use induction to prove this.

#### Base cases:
- For \( n = 2 \):
  \[
  \sigma = (2, 1)
  \]
  \[
  2 \cdot (-2)^0 + 1 \cdot (-2)^1 = 2 - 2 = 0
  \]
  This works.

- For \( n = 3 \):
  \[
  \sigma = (2, 3, 1)
  \]
  \[
  2 \cdot (-2)^0 + 3 \cdot (-2)^1 + 1 \cdot (-2)^2 = 2 - 6 + 4 = 0
  \]
  This works.

#### Inductive step:
Assume that for some \( n \equiv 0 \pmod{3} \) or \( n \equiv 2 \pmod{3} \), there exists a permutation \(\sigma\) that satisfies the condition. We need to show that it works for \( n+3 \).

Construct a new permutation \(\sigma_1\) for \( n+3 \) as follows:
\[
\sigma_1(1) = 2, \sigma_1(2) = 3, \sigma_1(n+3) = 1
\]
and for \( i \in \{3, \cdots, n+2\} \):
\[
\sigma_1(i) = \sigma(i-2) + 3
\]

Verify that this construction works:
\[
\sum_{i=1}^{n+3} \sigma_1(i)(-2)^{i-1} = 2 \cdot (-2)^0 + 3 \cdot (-2)^1 + \sum_{i=3}^{n+2} (\sigma(i-2) + 3)(-2)^{i-1} + 1 \cdot (-2)^{n+2}
\]
\[
= 2 - 6 + \sum_{i=1}^{n} \sigma(i)(-2)^{i+1} + 1 \cdot (-2)^{n+2}
\]
\[
= 2 - 6 + \sum_{i=1}^{n} \sigma(i)(-2)^{i+1} + (-2)^{n+2}
\]
Since \(\sum_{i=1}^{n} \sigma(i)(-2)^{i-1} = 0\), the sum simplifies to:
\[
2 - 6 + 0 + (-2)^{n+2} = 0
\]

Thus, the construction works for \( n+3 \).

### Conclusion
Combining the results from the lemmas, we conclude that the only \( n \) that work are those for which \( n \equiv 0 \pmod{3} \) or \( n \equiv 2 \pmod{3} \).

The final answer is \( \boxed{ n \equiv 0 \pmod{3} } \) or \( n \equiv 2 \pmod{3} \).