Celeste has an unlimited amount of each type of $n$ types of candy, numerated type 1, type 2, ... type n. Initially she takes $m>0$ candy pieces and places them in a row on a table. Then, she chooses one of the following operations (if available) and executes it:

$1.$ She eats a candy of type $k$, and in its position in the row she places one candy type $k-1$ followed by one candy type $k+1$ (we consider type $n+1$ to be type 1, and type 0 to be type $n$).

$2.$ She chooses two consecutive candies which are the same type, and eats them.

Find all positive integers $n$ for which Celeste can leave the table empty for any value of $m$ and any configuration of candies on the table.

$\textit{Proposed by Federico Bach and Santiago Rodriguez, Colombia}$
We need to determine for which positive integers \( n \) Celeste can always leave the table empty regardless of the initial number \( m \) and configuration of candies. We will analyze the problem by considering the operations and the effect they have on the candies.

1. **Operations Analysis**:
    - **Split Operation**: Replace a candy of type \( k \) with candies of type \( k-1 \) and \( k+1 \). Note that type \( n+1 \) is considered type 1, and type 0 is considered type \( n \).
    - **Take Operation**: Remove two consecutive candies of the same type.

2. **Case 1: \( n \equiv 0 \pmod{3} \)**:
    - Consider \( n = 3k \). We will show that it is impossible to empty the table for any initial configuration.
    - When \( m = 1 \), we start with one candy of some type \( a \). We can represent the types modulo 3, i.e., \( \{0, 1, 2\} \).
    - Each split operation changes the parity of the number of candies in each group (0, 1, 2 modulo 3). Specifically, if we split a candy of type \( k \), the parities of the groups \( k-1 \) and \( k+1 \) change.
    - The take operation does not change the parity of the number of candies in each group.
    - To empty the table, the number of candies in each group must be even. However, since the split operation changes the parity, it is impossible to achieve an even number of candies in each group starting from an odd number.

3. **Case 2: \( n \equiv 1 \pmod{3} \)**:
    - Consider \( n = 3k + 1 \). We will show that it is possible to empty the table for any initial configuration.
    - We can reduce the number of candies of any type by 1 without affecting the other types. This is done by repeatedly splitting and then taking pairs.
    - For example, if we start with a candy of type \( x \), we can split it to get \( x-1 \) and \( x+1 \). Repeating this process, we can eventually reduce the number of candies of type \( x \) to zero.
    - By carefully choosing the sequence of splits and takes, we can ensure that all candies are removed.

4. **Case 3: \( n \equiv 2 \pmod{3} \)**:
    - Consider \( n = 3k + 2 \). We will show that it is possible to empty the table for any initial configuration.
    - We can increase the number of candies of any type by 1 without affecting the other types. This is done by repeatedly splitting and then taking pairs.
    - For example, if we start with a candy of type \( x \), we can split it to get \( x-1 \) and \( x+1 \). Repeating this process, we can eventually increase the number of candies of type \( x \) to zero.
    - By carefully choosing the sequence of splits and takes, we can ensure that all candies are removed.

Combining these cases, we conclude that Celeste can leave the table empty for any initial configuration if and only if \( n \not\equiv 0 \pmod{3} \).

The final answer is \( \boxed{ n \not\equiv 0 \pmod{3} } \).