Describe which positive integers do not belong to the set
\[E = \left\{ \lfloor n+ \sqrt n +\frac 12 \rfloor | n \in \mathbb N\right\}.\]
To determine which positive integers do not belong to the set
\[E = \left\{ \lfloor n+ \sqrt n +\frac 12 \rfloor \mid n \in \mathbb{N} \right\},\]
we need to analyze the expression \(\lfloor n + \sqrt{n} + \frac{1}{2} \rfloor\).

1. **Expression Simplification**:
   Since \(n\) is a positive integer, we can rewrite the expression as:
   \[
   \lfloor n + \sqrt{n} + \frac{1}{2} \rfloor = n + \lfloor \sqrt{n} + \frac{1}{2} \rfloor.
   \]
   This is because the floor function \(\lfloor \cdot \rfloor\) applied to the sum of an integer and a non-integer part is the sum of the integer and the floor of the non-integer part.

2. **Case Analysis for \(n = t^2\)**:
   Let \(n = t^2\) where \(t\) is a positive integer. Then:
   \[
   \lfloor t^2 + \sqrt{t^2} + \frac{1}{2} \rfloor = \lfloor t^2 + t + \frac{1}{2} \rfloor.
   \]
   Since \(t\) is an integer, \(t + \frac{1}{2}\) is not an integer, and thus:
   \[
   \lfloor t^2 + t + \frac{1}{2} \rfloor = t^2 + t.
   \]

3. **General Case for \(n\) Between Two Squares**:
   Let \(x\) be the unique integer such that \(x^2 \leq n < (x+1)^2\). Then we can write \(n\) as:
   \[
   n = x^2 + k \quad \text{where} \quad 0 \leq k < 2x + 1.
   \]
   Now consider the expression:
   \[
   \lfloor n + \sqrt{n} + \frac{1}{2} \rfloor = \lfloor x^2 + k + \sqrt{x^2 + k} + \frac{1}{2} \rfloor.
   \]
   Since \(x^2 \leq n < (x+1)^2\), we have \(x \leq \sqrt{n} < x+1\). Therefore:
   \[
   x + \frac{1}{2} \leq \sqrt{n} + \frac{1}{2} < x + 1 + \frac{1}{2}.
   \]
   Adding \(n\) to these inequalities:
   \[
   n + x + \frac{1}{2} \leq n + \sqrt{n} + \frac{1}{2} < n + x + 1 + \frac{1}{2}.
   \]
   Taking the floor function:
   \[
   \lfloor n + x + \frac{1}{2} \rfloor \leq \lfloor n + \sqrt{n} + \frac{1}{2} \rfloor < \lfloor n + x + 1 + \frac{1}{2} \rfloor.
   \]
   Since \(n\) is an integer, \(\lfloor n + x + \frac{1}{2} \rfloor = n + x\) and \(\lfloor n + x + 1 + \frac{1}{2} \rfloor = n + x + 1\). Thus:
   \[
   n + x \leq \lfloor n + \sqrt{n} + \frac{1}{2} \rfloor < n + x + 1.
   \]
   Therefore:
   \[
   \lfloor n + \sqrt{n} + \frac{1}{2} \rfloor = n + x.
   \]

4. **Conclusion**:
   For \(n = t^2\), the expression \(\lfloor n + \sqrt{n} + \frac{1}{2} \rfloor\) equals \(t^2 + t\). Therefore, the set \(E\) misses exactly the squares of integers, as these values are not of the form \(n + x\) for \(n\) between two consecutive squares.

The final answer is the set \( \boxed{ E } \) misses exactly the squares of integers.