We are tasked with determining all integers $n \geq 2$ such that for any integers $a_1, a_2, \ldots, a_n$, whose sum is not divisible by $n$, there exists an index $1 \leq i \leq n$ such that none of the numbers

$$a_i, a_i + a_{i+1}, \ldots, a_i + a_{i+1} + \ldots + a_{i+n-1}$$

is divisible by $n$, with the circular definition $a_i = a_{i-n}$ when $i > n$.

To solve this problem, we will utilize properties of numbers and modular arithmetic.

### Step-by-Step Analysis:

1. Total Sum and Modular Arithmetic:

Considering the circular nature and divisibility, note: for each index $i$, the complete sum:

$$S_i = a_i + a_{i+1} + \ldots + a_{i+n-1}$$

If we assume $S = a_1 + a_2 + \ldots + a_n \equiv k \pmod{n}$ (where $k \neq 0$), then there must exist at least one index $i$ such that:

$$S_i \not\equiv 0 \pmod{n}$$

2. Condition for Prime Numbers:

If $n$ is a prime number, then the structure of the cyclic groups and the behavior under modular arithmetic facilitates that no full sum $S_i$ may default to a zero residue without contradicting $S \mod n \neq 0$ per problem condition.

3. Non-Prime Numbers:

Conversely, if $n$ is composite, there's potential to construct sequences where every partial sum $S_i \equiv 0 \pmod{n}$ due to factorization properties allowing divisions of full cycles into complete sub-cycles within the sequence subset.

### Conclusion:

Our primary result emerges from contradiction upon assumption, hence: for the described property to hold, the modulus $n$ must be prime. This way, there exists an index $i$ that satisfies the $n$-cyclic non-divisibility—able to subvert any aligning presence of zero residues across the complete rotation of terms.

Therefore, all integers $n \geq 2$ with the described property are precisely all prime numbers. The final answer is:
$$\boxed{\text{All prime numbers}}$$

Through this reasoning, it is confirmed that all and only prime numbers possess the trait of guaranteed non-zero residues in any set and rotation under restricted modulus sums, in compliance with the reference answer.