1. Define the sequences: Let $(x_n)_{n>0}$ and $(y_n)_{n>0}$ be sequences defined by $x_n = \frac{a_n}{n}$ and $y_n = \frac{b_n}{n}$ for all $n \geq 1$. We aim to show that $x_{1010} = y_{1009}$.

2. Transform the recurrence relations:
- For $(a_n)$, the recurrence relation is $a_{n+1} = \frac{2018}{n} a_n + a_{n-1}$.
- For $(b_n)$, the recurrence relation is $b_{n+1} = \frac{2020}{n} b_n + b_{n-1}$.

3. Express the generating functions:
- Let $T(x) = \sum_{n=1}^{\infty} x_n x^n$.
- From the recurrence relation for $a_n$, we have:
$$T'(x) - 1 = 2018 T(x) + x^2 T'(x).$$
- Rearrange and solve for $T(x)$:
$$T'(x) - 1 = 2018 T(x) + x^2 T'(x) \implies T'(x)(1 - x^2) = 2018 T(x) + 1.$$
- Let $P(x) = T(x) + \frac{1}{2018}$, then:
$$2018 P(x) = P'(x)(1 - x^2).$$

4. Solve the differential equation:
- Similarly, for $Q(x) = \sum_{n=1}^{\infty} y_n x^n + \frac{1}{2020}$, we get:
$$2020 Q(x) = Q'(x)(1 - x^2).$$
- Let $R(x)$ be the solution to $R(x) = R'(x)(1 - x^2)$ with $R(0) = 1$.

5. **Find the general solution for $R(x)$**:
- Solve the differential equation:
$$\frac{R'(x)}{R(x)} = \frac{1}{1 - x^2} \implies \int \frac{R'(x)}{R(x)} \, dx = \frac{1}{2} \int \left( \frac{1}{1 + x} + \frac{1}{1 - x} \right) \, dx.$$
- Integrate both sides:
$$\log(R(x)) = \frac{1}{2} (\log(1 + x) - \log(1 - x)) + c_1 \implies R(x) = c_2 (1 + x)^{1/2} (1 - x)^{-1/2}.$$
- Given $R(0) = 1$, we find $c_2 = 1$, so:
$$R(x) = (1 + x)^{1/2} (1 - x)^{-1/2}.$$

6. **Express $P(x)$ and $Q(x)$**:
- We have:
$$P(x) = \frac{R(x)^{2018}}{2018}, \quad Q(x) = \frac{R(x)^{2020}}{2020}.$$

7. Find the coefficients:
- For $x_{1010}$:
$$x_{1010} = [x^{1010}] P(x) = \frac{1}{2018} [x^{1010}] \left( (1 + x)^{1009} (1 - x)^{-1009} \right).$$
- For $y_{1009}$:
$$y_{1009} = [x^{1009}] Q(x) = \frac{1}{2020} [x^{1009}] \left( (1 + x)^{1010} (1 - x)^{-1010} \right).$$

8. Evaluate the coefficients:
- Using the binomial series expansion:
$$(1 + x)^{1009} (1 - x)^{-1009} = \sum_{k=0}^{1009} \binom{1009}{k} x^k \sum_{k=0}^{\infty} \binom{k + 1008}{1008} x^k.$$
- Similarly for $y_{1009}$:
$$(1 + x)^{1010} (1 - x)^{-1010} = \sum_{k=0}^{1010} \binom{1010}{k} x^k \sum_{k=0}^{\infty} \binom{k + 1009}{1009} x^k.$$

9. Compare the sums:
- We find:
$$x_{1010} = \sum_{k=0}^{1009} \frac{(2018 - k)!}{2 \cdot k! \cdot (1009 - k)! \cdot (1010 - k)!},$$
$$y_{1009} = \sum_{k=0}^{1009} \frac{(2018 - k)!}{2 \cdot k! \cdot (1010 - k)! \cdot (1009 - k)!}.$$
- Therefore, $x_{1010} = y_{1009}$.

The final answer is $\boxed{\frac{a_{1010}}{1010} = \frac{b_{1009}}{1009}}$