If $ n$ runs through all the positive integers, then $ f(n) \equal{} \left \lfloor n \plus{} \sqrt {3n} \plus{} \frac {1}{2} \right \rfloor$ runs through all positive integers skipping the terms of the sequence $ a_n \equal{} \left \lfloor \frac {n^2 \plus{} 2n}{3} \right \rfloor$.
1. **Define the sequences:**
   We are given two sequences:
   - \( f(n) = \left\lfloor n + \sqrt{3n} + \frac{1}{2} \right\rfloor \)
   - \( a_n = \left\lfloor \frac{n^2 + 2n}{3} \right\rfloor \)

2. **Objective:**
   We need to show that \( f(n) \) runs through all positive integers, skipping the terms of the sequence \( a_n \).

3. **Sequence \( b_n \):**
   Denote the sequence \( \{b_n\} \) as the sequence that runs through all positive integers skipping the terms of \( \{a_n\} \). We need to show that \( b_n = f(n) \).

4. **Inequality for \( b_n \):**
   Let \( b_n = k \). Then, \( k \) must lie between two consecutive terms of \( a_n \):
   \[
   a_m < k < a_{m+1}
   \]
   This implies that there are \( m \) terms of \( \{a_n\} \) and \( n \) terms of \( \{b_n\} \) in the set \(\{1, 2, \ldots, k\}\). Therefore, we have:
   \[
   k = n + m
   \]

5. **Bounds for \( k \):**
   Using the definition of \( a_n \), we have:
   \[
   \left\lfloor \frac{m(m + 2)}{3} \right\rfloor < n + m < \left\lfloor \frac{(m + 1)(m + 3)}{3} \right\rfloor
   \]
   This can be rewritten as:
   \[
   \frac{m(m + 2) - 2}{3} + 1 \leq \left\lfloor \frac{m(m + 2)}{3} \right\rfloor + 1 \leq n + m \leq \left\lfloor \frac{(m + 1)(m + 3)}{3} \right\rfloor - 1 \leq \frac{(m + 1)(m + 3)}{3} - 1
   \]

6. **Simplifying the bounds:**
   From the above inequalities, we get:
   \[
   \frac{m^2 + 2m - 2}{3} + 1 \leq n + m \leq \frac{m^2 + 4m + 3}{3} - 1
   \]
   Simplifying further:
   \[
   \frac{m^2 + 2m - 2 + 3}{3} \leq n + m \leq \frac{m^2 + 4m + 3 - 3}{3}
   \]
   \[
   \frac{m^2 + 2m + 1}{3} \leq n + m \leq \frac{m^2 + 4m}{3}
   \]

7. **Relating \( n \) and \( m \):**
   Rearranging the terms, we get:
   \[
   \frac{m^2 + 2m + 1}{3} - m \leq n \leq \frac{m^2 + 4m}{3} - m
   \]
   \[
   \frac{m^2 - m + 1}{3} \leq n \leq \frac{m^2 + m}{3}
   \]

8. **Bounding \( \sqrt{3n} \):**
   From the inequalities, we have:
   \[
   m^2 - m + 1 \leq 3n \leq m^2 + m
   \]
   Taking square roots:
   \[
   \sqrt{m^2 - m + 1} \leq \sqrt{3n} \leq \sqrt{m^2 + m}
   \]
   Approximating the square roots:
   \[
   m - \frac{1}{2} < \sqrt{3n} < m + \frac{1}{2}
   \]

9. **Floor function:**
   From the above, we get:
   \[
   m < \sqrt{3n} + \frac{1}{2} < m + 1
   \]
   Therefore:
   \[
   \left\lfloor \sqrt{3n} + \frac{1}{2} \right\rfloor = m
   \]

10. **Final expression for \( b_n \):**
    Substituting \( m \) back, we get:
    \[
    b_n = n + \left\lfloor \sqrt{3n} + \frac{1}{2} \right\rfloor
    \]
    This matches the given function \( f(n) \).

The final answer is \( \boxed{ f(n) = \left\lfloor n + \sqrt{3n} + \frac{1}{2} \right\rfloor } \).