Let $f_n$ be a sequence defined by $f_0=2020$ and 
\[f_{n+1} = \frac{f_n + 2020}{2020f_n + 1}\]
for all $n \geq 0$. Determine $f_{2020}$.
1. **Define the sequence and transformation:**
   Given the sequence \( f_n \) defined by:
   \[
   f_0 = 2020
   \]
   and
   \[
   f_{n+1} = \frac{f_n + 2020}{2020f_n + 1}
   \]
   for all \( n \geq 0 \).

2. **Introduce the hyperbolic tangent inverse function:**
   Let \( g_n = \tanh^{-1}(f_n) \). We need to express \( f_{n+1} \) in terms of \( g_n \).

3. **Use the addition formula for hyperbolic tangent:**
   Recall the addition formula for hyperbolic tangent:
   \[
   \tanh(a + b) = \frac{\tanh(a) + \tanh(b)}{1 + \tanh(a)\tanh(b)}
   \]
   Applying this to our sequence, we have:
   \[
   \tanh(g_{n+1}) = \tanh(g_n + \tanh^{-1}(2020))
   \]

4. **Simplify using the given sequence definition:**
   Since \( g_n = \tanh^{-1}(f_n) \), we can write:
   \[
   \tanh(g_{n+1}) = \tanh(g_n + \tanh^{-1}(2020))
   \]
   This implies:
   \[
   g_{n+1} = g_n + \tanh^{-1}(2020)
   \]

5. **Initial condition and recursive relation:**
   Given \( g_0 = \tanh^{-1}(2020) \), we can recursively find \( g_n \):
   \[
   g_1 = g_0 + \tanh^{-1}(2020) = 2\tanh^{-1}(2020)
   \]
   \[
   g_2 = g_1 + \tanh^{-1}(2020) = 3\tanh^{-1}(2020)
   \]
   Continuing this pattern, we get:
   \[
   g_n = (n+1)\tanh^{-1}(2020)
   \]

6. **Determine \( f_{2020} \):**
   For \( n = 2020 \):
   \[
   g_{2020} = 2021 \tanh^{-1}(2020)
   \]
   Therefore:
   \[
   f_{2020} = \tanh(g_{2020}) = \tanh(2021 \tanh^{-1}(2020))
   \]

Conclusion:
\[
f_{2020} = \boxed{\tanh(2021 \tanh^{-1}(2020))}
\]