Determine all possible values of integer $k$ for which there exist positive integers $a$ and $b$ such that $\dfrac{b+1}{a} + \dfrac{a+1}{b} = k$.
To determine all possible values of the integer \( k \) for which there exist positive integers \( a \) and \( b \) such that 

\[
\frac{b+1}{a} + \frac{a+1}{b} = k,
\]

we will analyze the equation step by step.

1. **Rewrite the equation:**
   \[
   \frac{b+1}{a} + \frac{a+1}{b} = k.
   \]

2. **Multiply both sides by \( ab \) to clear the denominators:**
   \[
   b(b+1) + a(a+1) = kab.
   \]

3. **Expand and simplify:**
   \[
   b^2 + b + a^2 + a = kab.
   \]

4. **Rearrange the equation:**
   \[
   a^2 + a + b^2 + b = kab.
   \]

5. **Consider the case \( a = b \):**
   \[
   \frac{b+1}{b} + \frac{b+1}{b} = k \implies 2 \left( \frac{b+1}{b} \right) = k \implies 2 \left( 1 + \frac{1}{b} \right) = k \implies k = 2 + \frac{2}{b}.
   \]
   Since \( k \) must be an integer, \( \frac{2}{b} \) must also be an integer. Therefore, \( b \) must be a divisor of 2. The positive divisors of 2 are 1 and 2.

   - If \( b = 1 \):
     \[
     k = 2 + \frac{2}{1} = 4.
     \]
   - If \( b = 2 \):
     \[
     k = 2 + \frac{2}{2} = 3.
     \]

6. **Consider the case \( a \neq b \):**
   Without loss of generality, assume \( a > b \). Let \( a = b + d \) where \( d \geq 1 \).

   Substitute \( a = b + d \) into the original equation:
   \[
   \frac{b+1}{b+d} + \frac{b+d+1}{b} = k.
   \]

7. **Analyze the expression:**
   \[
   \frac{b+1}{b+d} + \frac{b+d+1}{b}.
   \]

   - For \( d = 1 \):
     \[
     \frac{b+1}{b+1} + \frac{b+2}{b} = 1 + 1 + \frac{2}{b} = 2 + \frac{2}{b}.
     \]
     This is the same form as before, so \( k = 2 + \frac{2}{b} \). The possible values of \( k \) are still 3 and 4.

   - For \( d \geq 2 \):
     \[
     \frac{b+1}{b+d} + \frac{b+d+1}{b}.
     \]
     Since \( \frac{b+1}{b+d} < 1 \) and \( \frac{b+d+1}{b} > 1 \), the sum will be greater than 2 but less than 3. Therefore, no new integer values of \( k \) are possible.

Thus, the only possible integer values of \( k \) are 3 and 4.

The final answer is \( \boxed{3 \text{ and } 4} \).