Does the system of equation
\begin{align*}
\begin{cases}
x_1 + x_2 &= y_1 + y_2 + y_3 + y_4 \\
x_1^2 + x_2^2 &= y_1^2 + y_2^2 + y_3^2 + y_4^2 \\
x_1^3 + x_2^3 &= y_1^3 + y_2^3 + y_3^3 + y_4^3
\end{cases}
\end{align*}
admit a solution in integers such that the absolute value of each of these integers is greater than $2020$?
To determine if the given system of equations admits a solution in integers such that the absolute value of each of these integers is greater than $2020$, we can proceed as follows:

1. **Rewrite the system of equations:**
   \[
   \begin{cases}
   x_1 + x_2 &= y_1 + y_2 + y_3 + y_4 \\
   x_1^2 + x_2^2 &= y_1^2 + y_2^2 + y_3^2 + y_4^2 \\
   x_1^3 + x_2^3 &= y_1^3 + y_2^3 + y_3^3 + y_4^3
   \end{cases}
   \]

2. **Simplify the problem using symmetry:**
   Assume \( x_2 = -x_1 \), \( y_2 = -y_1 \), and \( y_4 = -y_3 \). This simplifies the equations as follows:
   \[
   \begin{cases}
   x_1 - x_1 &= y_1 - y_1 + y_3 - y_3 \\
   x_1^2 + x_1^2 &= y_1^2 + y_1^2 + y_3^2 + y_3^2 \\
   x_1^3 - x_1^3 &= y_1^3 - y_1^3 + y_3^3 - y_3^3
   \end{cases}
   \]
   Simplifying further, we get:
   \[
   \begin{cases}
   0 &= 0 \\
   2x_1^2 &= 2y_1^2 + 2y_3^2 \\
   0 &= 0
   \end{cases}
   \]

3. **Focus on the non-trivial equation:**
   The non-trivial equation is:
   \[
   x_1^2 = y_1^2 + y_3^2
   \]
   This equation can be satisfied by choosing \( x_1 \), \( y_1 \), and \( y_3 \) such that they form a Pythagorean triple. For example, if \( x_1 = 5 \), \( y_1 = 3 \), and \( y_3 = 4 \), then:
   \[
   5^2 = 3^2 + 4^2 \implies 25 = 9 + 16
   \]

4. **Scale the solution to satisfy the absolute value condition:**
   If \( (x_1, y_1, y_3) \) is a solution, then \( (Nx_1, Ny_1, Ny_3) \) is also a solution for any integer \( N \). To ensure the absolute value of each integer is greater than \( 2020 \), choose \( N \) such that:
   \[
   N \cdot \min(|x_1|, |y_1|, |y_3|) > 2020
   \]
   For the example above, \( \min(|5|, |3|, |4|) = 3 \), so choose \( N \) such that:
   \[
   3N > 2020 \implies N > \frac{2020}{3} \approx 673.33
   \]
   Therefore, \( N = 674 \) is a suitable choice.

5. **Construct the final solution:**
   Using \( N = 674 \), the solution is:
   \[
   x_1 = 674 \cdot 5 = 3370, \quad x_2 = -3370
   \]
   \[
   y_1 = 674 \cdot 3 = 2022, \quad y_2 = -2022
   \]
   \[
   y_3 = 674 \cdot 4 = 2696, \quad y_4 = -2696
   \]

The final answer is \(\boxed{\text{True}}\).