Find all pairs of positive integers $(n, k)$ such that all sufficiently large odd positive integers $m$ are representable as $$m=a_1^{n^2}+a_2^{(n+1)^2}+\ldots+a_k^{(n+k-1)^2}+a_{k+1}^{(n+k)^2}$$ for some non-negative integers $a_1, a_2, \ldots, a_{k+1}$.
1. We start by considering the given equation for representing a sufficiently large odd positive integer \( m \):
   \[
   m = a_1^{n^2} + a_2^{(n+1)^2} + \ldots + a_k^{(n+k-1)^2} + a_{k+1}^{(n+k)^2}
   \]
   where \( a_1, a_2, \ldots, a_{k+1} \) are non-negative integers.

2. To determine the number of possible values for \( a_1 \) that can represent \( m \leq N \), we note that \( a_1 \) must satisfy \( a_1^{n^2} \leq N \). Therefore, the number of possible values for \( a_1 \) is approximately \( N^{\frac{1}{n^2}} \).

3. Similarly, the number of possible values for \( a_2 \) is approximately \( N^{\frac{1}{(n+1)^2}} \), and so on. Thus, the total number of possible combinations of \( a_1, a_2, \ldots, a_{k+1} \) is at most:
   \[
   N^{\frac{1}{n^2}} \times N^{\frac{1}{(n+1)^2}} \times \ldots \times N^{\frac{1}{(n+k)^2}} = N^{\sum_{i=0}^{k} \frac{1}{(n+i)^2}}
   \]

4. For \( m \) to be representable as \( m = a_1^{n^2} + a_2^{(n+1)^2} + \ldots + a_k^{(n+k-1)^2} + a_{k+1}^{(n+k)^2} \) for all sufficiently large odd \( m \), the exponent in the expression \( N^{\sum_{i=0}^{k} \frac{1}{(n+i)^2}} \) must be at least 1. Therefore, we require:
   \[
   \sum_{i=0}^{k} \frac{1}{(n+i)^2} \geq 1
   \]

5. We now analyze the sum \( \sum_{i=0}^{k} \frac{1}{(n+i)^2} \). If \( n \geq 2 \), the sum of the series \( \sum_{i=0}^{\infty} \frac{1}{(n+i)^2} \) converges to a value less than 1, specifically:
   \[
   \sum_{i=0}^{\infty} \frac{1}{(n+i)^2} < \sum_{i=1}^{\infty} \frac{1}{i^2} = \frac{\pi^2}{6} \approx 1.644
   \]
   Since \( \frac{\pi^2}{6} - 1 < 1 \), it implies that for any finite \( k \), the sum \( \sum_{i=0}^{k} \frac{1}{(n+i)^2} \) will be even smaller, and thus cannot reach 1 if \( n \geq 2 \).

6. Therefore, the only possibility is \( n = 1 \). For \( n = 1 \), the sum becomes:
   \[
   \sum_{i=0}^{k} \frac{1}{(1+i)^2} = \sum_{i=1}^{k+1} \frac{1}{i^2}
   \]
   This sum can be made arbitrarily large by choosing a sufficiently large \( k \). Hence, for \( n = 1 \), any \( k \) will work.

7. To verify, if \( n = 1 \), we can always represent \( m \) by choosing \( a_1 = m \) and setting the rest of the \( a_i \)'s to zero, ensuring that any \( k \) works.

Conclusion:
The only pairs \((n, k)\) that satisfy the condition are \((1, k)\) for any positive integer \( k \).

The final answer is \( \boxed{ (1, k) } \) for any positive integer \( k \).