1. We start with the given condition that for $n \in \mathbb{N}$, the function $f$ is non-negative and defined on the set $\{1, 2, \ldots, n\}$ such that $f(i) \geq f(j)$ for $i > j$. Additionally, we have the equation:
$$\sum_{i=1}^{n} (i + f(i)) = 2023$$

2. We can separate the sum into two parts:
$$\sum_{i=1}^{n} i + \sum_{i=1}^{n} f(i) = 2023$$

3. The first sum is the sum of the first $n$ natural numbers, which is given by:
$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$

4. Substituting this into the equation, we get:
$$\frac{n(n+1)}{2} + \sum_{i=1}^{n} f(i) = 2023$$

5. Solving for $\sum_{i=1}^{n} f(i)$, we have:
$$\sum_{i=1}^{n} f(i) = 2023 - \frac{n(n+1)}{2}$$

6. To minimize $\sum_{i=1}^{n} f(i)$, we need to maximize $\frac{n(n+1)}{2}$ while ensuring that $\sum_{i=1}^{n} f(i)$ remains non-negative. This means:
$$2023 - \frac{n(n+1)}{2} \geq 0$$

7. Solving the inequality:
$$2023 \geq \frac{n(n+1)}{2}$$
$$4046 \geq n(n+1)$$

8. We need to find the largest integer $n$ such that $n(n+1) \leq 4046$. Testing values, we find:
$$n = 63 \implies 63 \times 64 = 4032 \quad (\text{valid})$$
$$n = 64 \implies 64 \times 65 = 4160 \quad (\text{invalid})$$

Therefore, the maximum $n$ is 63.

9. Substituting $n = 63$ back into the equation for $\sum_{i=1}^{n} f(i)$:
$$\sum_{i=1}^{n} f(i) = 2023 - \frac{63 \times 64}{2} = 2023 - 2016 = 7$$

10. We need to count the number of non-negative integer solutions to the equation:
$$\sum_{i=1}^{63} f(i) = 7$$
where $f(i) \geq f(j)$ for $i > j$. This is equivalent to finding the number of partitions of 7 into at most 63 parts, which is the same as the number of partitions of 7.

11. The number of partitions of 7 is given by:
$$p(7) = 15$$

The final answer is $\boxed{15}$.