1. We are given that $a, b, c, d, (a + b + c + 18 + d), (a + b + c + 18 - d), (b + c),$ and $(c + d)$ are distinct prime numbers, and $a + b + c = 2010$. Additionally, $a, b, c, d \neq 3$ and $d \leq 50$.

2. From the conditions, we have:
$$a + b + c + 18 + d = 2028 + d \quad \text{(prime number)}$$
$$a + b + c + 18 - d = 2028 - d \quad \text{(prime number)}$$
Since both $2028 + d$ and $2028 - d$ must be prime, $d$ must be odd (as adding or subtracting an even number from an even number results in an even number, which cannot be prime except for 2, but $d \neq 3$).

3. Next, consider $c + d$. Since $c + d$ is a prime number greater than 2, it must be odd. Therefore, $c$ must be even. The only even prime number is 2, so $c = 2$.

4. Substituting $c = 2$ into $a + b + c = 2010$, we get:
$$a + b + 2 = 2010 \implies a + b = 2008$$

5. Now, we need to ensure that $2028 + d$ and $2028 - d$ are prime numbers. Additionally, $d \leq 50$ and $d$ must be such that both $2028 + d$ and $2028 - d$ are primes.

6. We also need to ensure that $d$ and $d + 2$ are both primes. The possible values of $d$ that satisfy these conditions are $d \in \{5, 11, 29, 41\}$.

7. Checking these values:
- For $d = 5$:
$$2028 + 5 = 2033 \quad (\text{not prime})$$
$$2028 - 5 = 2023 \quad (\text{prime})$$
- For $d = 11$:
$$2028 + 11 = 2039 \quad (\text{prime})$$
$$2028 - 11 = 2017 \quad (\text{prime})$$
- For $d = 29$:
$$2028 + 29 = 2057 \quad (\text{prime})$$
$$2028 - 29 = 1999 \quad (\text{prime})$$
- For $d = 41$:
$$2028 + 41 = 2069 \quad (\text{prime})$$
$$2028 - 41 = 1987 \quad (\text{prime})$$

8. Among these values, $d = 41$ gives the highest value for $2028 + d$. Therefore, the maximum value of the difference between two of these prime numbers is:
$$(a + b + c + 18 + d) - c = (2028 + d) - 2 = 2026 + d$$
Substituting $d = 41$:
$$2026 + 41 = 2067$$

The final answer is $\boxed{2067}$