To determine the natural numbers that cannot be written as $\lfloor n + \sqrt{n} + \frac{1}{2} \rfloor$ for any $n \in \mathbb{N}$, we will analyze the function $f(n) = n + \sqrt{n} + \frac{1}{2}$ and its properties.

1. Define the function and its floor value:
$$f(n) = n + \sqrt{n} + \frac{1}{2}$$
We are interested in the floor value:
$$\lfloor f(n) \rfloor = \lfloor n + \sqrt{n} + \frac{1}{2} \rfloor$$

2. Consider the function at consecutive integers:
$$f(n) = n + \sqrt{n} + \frac{1}{2}$$
$$f(n+1) = (n+1) + \sqrt{n+1} + \frac{1}{2}$$

3. **Calculate the difference between $f(n+1)$ and $f(n)$:**
$$f(n+1) - f(n) = (n+1 + \sqrt{n+1} + \frac{1}{2}) - (n + \sqrt{n} + \frac{1}{2})$$
$$f(n+1) - f(n) = 1 + \sqrt{n+1} - \sqrt{n}$$

4. Analyze the range of the difference:
Since $\sqrt{n+1} - \sqrt{n}$ is always less than 1 for all $n \in \mathbb{N}$:
$$1 < f(n+1) - f(n) < 2$$

5. Determine the conditions for skipping a number:
For a natural number $k$ to be skipped, it must lie between $\lfloor f(n) \rfloor$ and $\lfloor f(n+1) \rfloor$ without being equal to either. This happens if:
$$\lfloor f(n) \rfloor + 1 = \lfloor f(n+1) \rfloor$$
This implies:
$$\lfloor n + \sqrt{n} + \frac{1}{2} \rfloor + 1 = \lfloor n+1 + \sqrt{n+1} + \frac{1}{2} \rfloor$$

6. Identify the skipped numbers:
The skipped numbers occur when:
$$\{\sqrt{n}\} < \frac{1}{2} \quad \text{and} \quad \{\sqrt{n+1}\} > \frac{1}{2}$$
This is true when $n = k^2 + k$ for some integer $k$.

7. Verify the skipped numbers:
For $n = k^2 + k$:
$$f(k^2 + k) = k^2 + k + \sqrt{k^2 + k} + \frac{1}{2}$$
$$f(k^2 + k + 1) = k^2 + k + 1 + \sqrt{k^2 + k + 1} + \frac{1}{2}$$
The difference:
$$f(k^2 + k + 1) - f(k^2 + k) = 1 + \sqrt{k^2 + k + 1} - \sqrt{k^2 + k}$$
Since $\sqrt{k^2 + k + 1} - \sqrt{k^2 + k} < 1$, the floor values differ by 2:
$$\lfloor f(k^2 + k + 1) \rfloor - \lfloor f(k^2 + k) \rfloor = 2$$
The skipped number is:
$$\lfloor f(k^2 + k) \rfloor + 1 = (k+1)^2$$

8. Conclusion:
All and only perfect squares are skipped.

The final answer is all perfect squares.