Given two positive integers $n$ and $k$, we say that $k$ is [i]$n$-ergetic[/i] if:
However the elements of $M=\{1,2,\ldots, k\}$ are coloured in red and green, there exist $n$ not necessarily distinct integers of the same colour whose sum is again an element of $M$ of the same colour. For each positive integer $n$, determine the least $n$-ergetic integer, if it exists.
1. **Claim**: The minimum value of \( M \) that is \( n \)-ergetic is \( n^2 + n - 1 \).

2. **Base Case**: When \( n = 1 \), the claim is obvious because any integer \( k \geq 1 \) is trivially \( 1 \)-ergetic.

3. **Inductive Step**: Assume \( n > 1 \). We need to show that \( M = n^2 + n - 1 \) is \( n \)-ergetic.

4. **Construction**: Color the elements of \( M = \{1, 2, \ldots, n^2 + n - 1\} \) as follows:
   - Color \( 1 \) to \( n-1 \) red.
   - Color \( n \) to \( n^2 - 1 \) green.
   - Color \( n^2 \) to \( n^2 + n - 2 \) red.

5. **Verification**: Check that \( n^2 + n - 2 \) is not \( n \)-ergetic:
   - Suppose we have colored \( 1, 2, \ldots, n^2 + n - 2 \) such that no \( n \) integers of the same color sum to another integer of the same color.
   - Without loss of generality, assume \( 1 \) is red.
   - Let \( k \) be the largest integer such that \( 1, 2, \ldots, k \) are red. We claim \( k = 1 \).

6. **Proof by Contradiction**:
   - Since \( 1 \) added \( n \) times yields \( n \), \( n \) must be green. Thus, \( k < n \).
   - Each number in the interval \([n, nk]\) can be written as the sum of \( n \) integers from \([1, k]\), so these numbers are green.
   - Each number in the interval \([n^2, n^2 + n - 1]\) can be written as the sum of \( n \) integers from \([n, nk]\), so these numbers are red.
   - But \( n^2 \) along with \( n-1 \) copies of \( 1 \) contradicts the statement since \( n^2 + n - 1 \) is red. Hence, \( k = 1 \).

7. **Further Analysis**:
   - It follows that \( 2 \) is green.
   - Suppose \( u \) is the largest integer such that \([2, u]\) is green. We claim \( u = 2 \).
   - Each number in the interval \([2n, un]\) is red. If \( 2n + n - 1 \leq un \), then \( 2n \) and \( n-1 \) copies of \( n \) contradict our assumption. Therefore, \( u = 2 \).

8. **Conclusion**:
   - We have shown that \( 1, 2, 3 \) are red, green, and red respectively.
   - \( 3n - 2 \) can be written as the sum of \( n-1 \) copies of \( 3 \) and a single \( 1 \), so \( 3n - 2 \) must be green.
   - \( 3n - 2 \) can also be written as the sum of \( n-1 \) copies of \( 2 \) and a single \( n \) (which is green). This is the final contradiction showing that \( M = n^2 + n - 1 \) is indeed \( n \)-ergetic.

\(\blacksquare\)

The final answer is \( \boxed{ n^2 + n - 1 } \)