To determine all natural numbers $n$ for which there exists a permutation $(a_1, a_2, \ldots, a_n)$ of the numbers $0, 1, \ldots, n-1$ such that, if $b_i$ is the remainder of $a_1 + a_2 + \cdots + a_i$ upon division by $n$ for $i = 1, \ldots, n$, then $(b_1, b_2, \ldots, b_n)$ is also a permutation of $0, 1, \ldots, n-1$, we proceed as follows:

1. **Expression for $b_i$:**
$$b_i \equiv a_1 + a_2 + \cdots + a_i \pmod{n}$$
We need $(b_1, b_2, \ldots, b_n)$ to be a permutation of $0, 1, \ldots, n-1$.

2. **Sum of $b_i$:**
$$\sum_{i=1}^n b_i \equiv \sum_{i=1}^n (a_1 + a_2 + \cdots + a_i) \pmod{n}$$
This can be rewritten as:
$$\sum_{i=1}^n b_i \equiv n \cdot a_1 + (n-1) \cdot a_2 + \cdots + 2 \cdot a_{n-1} + a_n \pmod{n}$$

3. **Considering $n = 3$:**
For $n = 3$, we need to check if there exists a permutation $(a_1, a_2, a_3)$ of $0, 1, 2$ such that $(b_1, b_2, b_3)$ is a permutation of $0, 1, 2$.

4. General Case Analysis:
- If $n$ is even, then $n$ must be a prime number. If $n$ is composite, then at least two of the $b_i$'s will be equal, which contradicts the requirement that $(b_1, b_2, \ldots, b_n)$ is a permutation of $0, 1, \ldots, n-1$.

5. Constructing the Permutation:
- For $n$ to be a prime number, we can construct the permutation $(a_1, a_2, \ldots, a_n)$ such that $(b_1, b_2, \ldots, b_n)$ is a permutation of $0, 1, \ldots, n-1$.

6. **Example for $n = 5$:**
- Consider $n = 5$. We can take $(a_1, a_2, a_3, a_4, a_5) = (0, 4, 1, 3, 2)$.
- Then, $b_i$ will be:
$$b_1 = 0, \quad b_2 = 4, \quad b_3 = 0 + 4 + 1 = 5 \equiv 0 \pmod{5}, \quad b_4 = 0 + 4 + 1 + 3 = 8 \equiv 3 \pmod{5}, \quad b_5 = 0 + 4 + 1 + 3 + 2 = 10 \equiv 0 \pmod{5}$$
- This does not work, so we need to find another permutation.

7. Conclusion:
- The only natural numbers $n$ for which there exists such a permutation are prime numbers.

The final answer is $\boxed{ n }$ must be a prime number.