Natural numbers $k, l,p$ and $q$ are such that if $a$ and $b$ are roots of $x^2 - kx + l = 0$ then $a +\frac1b$ and $b + \frac1a$ are the roots of $x^2 -px + q = 0$. What is the sum of all possible values of $q$?
1. Assume the roots of the equation \(x^2 - kx + l = 0\) are \(\alpha\) and \(\beta\). By Vieta's formulas, we have:
   \[
   \alpha + \beta = k \quad \text{and} \quad \alpha \beta = l
   \]

2. We need to find an equation whose roots are \(\alpha + \frac{1}{\beta}\) and \(\beta + \frac{1}{\alpha}\). The sum of these roots is:
   \[
   \left(\alpha + \frac{1}{\beta}\right) + \left(\beta + \frac{1}{\alpha}\right) = \alpha + \beta + \frac{1}{\alpha} + \frac{1}{\beta}
   \]
   Using the relationships from Vieta's formulas, we can rewrite this as:
   \[
   \alpha + \beta + \frac{1}{\alpha} + \frac{1}{\beta} = k + \frac{1}{\alpha} + \frac{1}{\beta}
   \]
   Since \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}\), we get:
   \[
   k + \frac{\alpha + \beta}{\alpha \beta} = k + \frac{k}{l}
   \]
   Therefore, the sum of the roots is:
   \[
   k + \frac{k}{l} = \frac{k(l + 1)}{l}
   \]

3. The product of the roots \(\alpha + \frac{1}{\beta}\) and \(\beta + \frac{1}{\alpha}\) is:
   \[
   \left(\alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\right) = \alpha \beta + \alpha \cdot \frac{1}{\alpha} + \beta \cdot \frac{1}{\beta} + \frac{1}{\alpha \beta}
   \]
   Simplifying, we get:
   \[
   \alpha \beta + 1 + 1 + \frac{1}{\alpha \beta} = l + 2 + \frac{1}{l}
   \]
   Therefore, the product of the roots is:
   \[
   l + 2 + \frac{1}{l} = \frac{(l + 1)^2}{l}
   \]

4. We now have the quadratic equation with roots \(\alpha + \frac{1}{\beta}\) and \(\beta + \frac{1}{\alpha}\) as:
   \[
   x^2 - \left(\frac{k(l + 1)}{l}\right)x + \frac{(l + 1)^2}{l} = 0
   \]

5. Comparing this with the given equation \(x^2 - px + q = 0\), we get:
   \[
   p = \frac{k(l + 1)}{l} \quad \text{and} \quad q = \frac{(l + 1)^2}{l}
   \]

6. We need to find the sum of all possible values of \(q\). Since \(k, l, p,\) and \(q\) are natural numbers, we need \(l\) to be a natural number that makes \(q\) a natural number. Let's check the possible values of \(l\):

   - For \(l = 1\):
     \[
     q = \frac{(1 + 1)^2}{1} = \frac{4}{1} = 4
     \]

   - For \(l = 2\):
     \[
     q = \frac{(2 + 1)^2}{2} = \frac{9}{2} = 4.5 \quad \text{(not a natural number)}
     \]

   - For \(l = 3\):
     \[
     q = \frac{(3 + 1)^2}{3} = \frac{16}{3} \quad \text{(not a natural number)}
     \]

   - For \(l = 4\):
     \[
     q = \frac{(4 + 1)^2}{4} = \frac{25}{4} \quad \text{(not a natural number)}
     \]

   - For \(l = 5\):
     \[
     q = \frac{(5 + 1)^2}{5} = \frac{36}{5} \quad \text{(not a natural number)}
     \]

   From this, we see that the only natural number value for \(q\) is when \(l = 1\), giving \(q = 4\).

The final answer is \(\boxed{4}\)