Given the problem, we are tasked with finding all integers $n \geq 2$ such that there exists a permutation $x_0, x_1, \ldots, x_{n-1}$ of the numbers $0, 1, \ldots, n-1$ with the property that the $n$ numbers

$$x_0, \quad x_0 + x_1, \quad \ldots, \quad x_0 + x_1 + \ldots + x_{n-1}$$

are pairwise distinct modulo $n$.

To solve this, consider the sequence $S_k = x_0 + x_1 + \ldots + x_k$ for $0 \leq k < n$. We need $S_0, S_1, \ldots, S_{n-1}$ to be distinct modulo $n$.

### Step 1: Permutation and Constraints

Since $x_0, x_1, \ldots, x_{n-1}$ is a permutation of $0, 1, \ldots, n-1$, we have:

$$x_0 + x_1 + \ldots + x_{n-1} \equiv 0 + 1 + \ldots + (n-1) \equiv \frac{n(n-1)}{2} \pmod{n}$$

This reduces to:

$$x_0 + x_1 + \ldots + x_{n-1} \equiv 0 \pmod{n}$$

Thus, $S_{n-1} \equiv 0 \pmod{n}$.

### Step 2: Distinctness Condition

For $S_0, S_1, \ldots, S_{n-1}$ to be pairwise distinct modulo $n$, we need:

$$S_{k_1} \not\equiv S_{k_2} \pmod{n} \quad \text{for} \quad 0 \leq k_1 < k_2 < n$$

### Step 3: Consider Special Cases

- **Case $n = 2$:** The permutation $(0, 1)$ works because:

$$S_0 = 0, \quad S_1 = 0 + 1 = 1$$

Both are distinct modulo 2.

- **Even $n \geq 2$:** Consider the permutation that results in distinct cumulative sums. By constructing carefully, we can ensure the sums are distinct modulo $n$.

- **Odd $n$:** If $n$ is odd, consider:

$$S_k \equiv S_{k'} \pmod{n} \quad \text{for some } k < k' \text{ since } n \mid \frac{n(n-1)}{2}$$

Since $\frac{n(n-1)}{2}$ is an integer and $n$ odd, this leads to a contradiction of the distinct condition.

### Conclusion

Thus, the integer $n$ must be even for the condition to hold. Therefore, the solution is:

$$\boxed{\text{even integers } n \geq 2}$$

This concludes that all even integers greater than or equal to 2 satisfy the given property in the problem.