Let $f : \mathbb{N} \to \mathbb{N}$ be a function such that the following conditions hold:

$\qquad\ (1) \; f(1) = 1.$

$\qquad\ (2) \; \dfrac{(x + y)}{2} < f(x + y) \le f(x) + f(y) \; \forall \; x, y \in \mathbb{N}.$

$\qquad\ (3) \; f(4n + 1) < 2f(2n + 1) \; \forall \; n \ge 0.$

$\qquad\ (4) \; f(4n + 3) \le 2f(2n + 1) \; \forall \; n \ge 0.$


Find the sum of all possible values of $f(2023)$.
1. **Base Case Verification:**
   - Given \( f(1) = 1 \).
   - We need to verify the function for small values to establish a pattern.
   - For \( x = 1 \) and \( y = 1 \):
     \[
     \frac{1 + 1}{2} < f(2) \le f(1) + f(1) \implies 1 < f(2) \le 2
     \]
     Since \( f(2) \) must be a natural number, \( f(2) = 2 \).

2. **Verification for \( f(3) \):**
   - For \( x = 1 \) and \( y = 2 \):
     \[
     \frac{1 + 2}{2} < f(3) \le f(1) + f(2) \implies 1.5 < f(3) \le 3
     \]
     Also, from condition (4):
     \[
     f(3) \le 2f(1) \implies f(3) \le 2
     \]
     Combining these, we get \( f(3) = 2 \).

3. **Verification for \( f(4) \):**
   - For \( x = 2 \) and \( y = 2 \):
     \[
     \frac{2 + 2}{2} < f(4) \le f(2) + f(2) \implies 2 < f(4) \le 4
     \]
     Also, from condition (3):
     \[
     f(4n + 1) < 2f(2n + 1) \implies f(5) < 2f(3) \implies f(5) < 4
     \]
     Since \( f(5) \) must be a natural number and \( f(5) \le 3 \), we get \( f(5) = 3 \).

4. **Verification for \( f(6) \):**
   - For \( x = 3 \) and \( y = 3 \):
     \[
     \frac{3 + 3}{2} < f(6) \le f(3) + f(3) \implies 3 < f(6) \le 4
     \]
     Since \( f(6) \) must be a natural number, \( f(6) = 4 \).

5. **Inductive Hypothesis:**
   - Assume \( f(2n) = n + 1 \) and \( f(2n + 1) = n + 1 \) for some \( n \).

6. **Inductive Step:**
   - For \( f(2n + 2) \):
     \[
     \frac{2n + 2}{2} < f(2n + 2) \le f(2n + 1) + f(1) \implies n + 1 < f(2n + 2) \le n + 2
     \]
     Since \( f(2n + 2) \) must be a natural number, \( f(2n + 2) = n + 2 \).

   - For \( f(2n + 3) \):
     \[
     \frac{2n + 3}{2} < f(2n + 3) \le f(2n + 1) + f(2) \implies n + 1.5 < f(2n + 3) \le n + 3
     \]
     Also, from condition (4):
     \[
     f(4n + 3) \le 2f(2n + 1) \implies f(2n + 3) \le 2(n + 1)
     \]
     Since \( f(2n + 3) \) must be a natural number, \( f(2n + 3) = n + 2 \).

7. **Conclusion:**
   - By induction, \( f(n) = \left\lfloor \frac{n}{2} \right\rfloor + 1 \).

8. **Finding \( f(2023) \):**
   - Using the formula:
     \[
     f(2023) = \left\lfloor \frac{2023}{2} \right\rfloor + 1 = 1011 + 1 = 1012
     \]

The final answer is \( \boxed{1012} \)