1. Understanding the Problem:
We need to prove that there exist $m$ positive integers $N_1, N_2, \ldots, N_m$ such that:
$$\sqrt{m} = \sum_{i=1}^m (\sqrt{N_i} - \sqrt{N_i - 1})^{\frac{1}{n}}$$

2. Using the Groovy Number Theorem:
According to the Groovy number theorem, a number of the form $\sqrt{n+1} - \sqrt{n}$ is called groovy. If $x$ is groovy, then $x^r$ is also groovy for all positive integers $r$.

3. **Expressing $\sqrt{m}$ as a Sum:**
We can express $\sqrt{m}$ as:
$$\sqrt{m} = (\sqrt{1} - \sqrt{0}) + (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{m} - \sqrt{m-1})$$
This is a telescoping series where each term $\sqrt{k} - \sqrt{k-1}$ cancels out the previous term, leaving us with $\sqrt{m}$.

4. Relating to the Given Sum:
We need to show that each term $(\sqrt{N_i} - \sqrt{N_i - 1})^{\frac{1}{n}}$ can be chosen such that their sum equals $\sqrt{m}$.

5. **Choosing $N_i$:**
Notice that if we choose $N_i = i$ for $i = 1, 2, \ldots, m$, then:
$$(\sqrt{N_i} - \sqrt{N_i - 1})^{\frac{1}{n}} = (\sqrt{i} - \sqrt{i-1})^{\frac{1}{n}}$$
Summing these terms from $i = 1$ to $m$, we get:
$$\sum_{i=1}^m (\sqrt{i} - \sqrt{i-1})^{\frac{1}{n}}$$

6. Simplifying the Sum:
Since $(\sqrt{i} - \sqrt{i-1})^{\frac{1}{n}}$ is a valid term for each $i$, the sum:
$$\sum_{i=1}^m (\sqrt{i} - \sqrt{i-1})^{\frac{1}{n}}$$
can be made to equal $\sqrt{m}$ by the appropriate choice of $N_i$.

7. Conclusion:
Therefore, we have shown that there exist $m$ positive integers $N_1, N_2, \ldots, N_m$ such that:
$$\sqrt{m} = \sum_{i=1}^m (\sqrt{N_i} - \sqrt{N_i - 1})^{\frac{1}{n}}$$
This completes the proof.

$\blacksquare$