1. Define the sets and functions:
- Let $G(m,n) = \{ x \in \mathbb{N} \mid m \le x \le n; \; x \text{ is a good number} \}$
- Let $g(m,n) = |G(m,n)|$
- Let $B(m,n) = \{ y \in \mathbb{N} \mid m \le y \le n; \; y \text{ is a bad number} \}$
- Let $b(m,n) = |B(m,n)|$
- We need to find $d = g(1,2023) - b(1,2023)$

2. Break down the range:
- $g(1,2023) = g(1,1999) + g(2000,2023)$
- $b(1,2023) = b(1,1999) + b(2000,2023)$

3. **Prove $g(1,1999) = b(1,1999)$:**
- Let $U = \{1, 2, \dots, 1998, 1999\}$
- For $u \in U$, represent $u$ as $u = u_3 \cdot 10^3 + u_2 \cdot 10^2 + u_1 \cdot 10 + u_0$ where $u_3 \in \{0,1\}$ and $u_2, u_1, u_0 \in \{0,1,\dots,9\}$
- Define the function $T: U \to U$ such that:
$$v_k = \begin{cases} u_k & \text{if } u_k \in \{0,1,4,5,6,7,8,9\} \\ 3 & \text{if } u_k = 2 \\ 2 & \text{if } u_k = 3 \end{cases}$$
for $k \in \{0,1,2,3\}$

4. **Properties of $T$:**
- $T(T(u)) = u$ for all $u \in U$, hence $T$ is injective.
- If $u \in G(1,1999)$, then $T(u) \in B(1,1999)$, implying $g(1,1999) \le b(1,1999)$
- If $u \in B(1,1999)$, then $T(u) \in G(1,1999)$, implying $b(1,1999) \le g(1,1999)$

5. Conclusion from properties:
- From the above properties, we get $g(1,1999) = b(1,1999)$

6. **Calculate $g(2000,2023)$ and $b(2000,2023)$:**
- The set $V = [2000,2023] \cap \mathbb{N}$ contains 24 elements.
- Numbers 2003 and 2013 are neither good nor bad.
- Therefore, $V \setminus \{2003,2013\} \subset G(2000,2023)$
- Hence, $g(2000,2023) = 22$ and $b(2000,2023) = 0$

7. Final calculation:
- $d = g(2000,2023) - b(2000,2023) = 22 - 0 = 22$

The final answer is $\boxed{22}$