For each positive integer $k$, denote by $P (k)$ the number of all positive integers $4k$-digit numbers which can be composed of the digits $2, 0$ and which are divisible by the number $2 020$. Prove the inequality $$P (k)  \ge  \binom{2k - 1}{k}^2$$ and determine all $k$ for which equality occurs.

(Note: A positive integer cannot begin with a digit of $0$.)

(Jaromir Simsa)
1. **Define the problem and constraints:**
   For each positive integer \( k \), we need to find the number of \( 4k \)-digit numbers composed of the digits \( 2 \) and \( 0 \) that are divisible by \( 2020 \). We denote this number by \( P(k) \). We need to prove the inequality:
   \[
   P(k) \geq \binom{2k-1}{k}^2
   \]
   and determine all \( k \) for which equality holds.

2. **Analyze the divisibility conditions:**
   Since \( 2020 = 2^2 \cdot 5 \cdot 101 \), a number \( m \) must be divisible by \( 4 \), \( 5 \), and \( 101 \).

3. **Divisibility by 4 and 5:**
   - For \( m \) to be divisible by \( 4 \), the last two digits must be \( 00 \).
   - For \( m \) to be divisible by \( 5 \), the last digit must be \( 0 \). Since the last two digits are \( 00 \), both conditions are satisfied.

4. **Divisibility by 101:**
   - We need to consider the sum of the digits in positions that are congruent modulo 4. Let:
     \[
     A = \sum_{i \equiv 1 \pmod{4}} a_i, \quad B = \sum_{i \equiv 2 \pmod{4}} a_i, \quad C = \sum_{i \equiv 3 \pmod{4}} a_i, \quad D = \sum_{i \equiv 0 \pmod{4}} a_i
     \]
   - Using the properties of powers of 10 modulo 101, we have:
     \[
     10^{4k-i} \equiv \begin{cases} 
     -10 & \text{if } i \equiv 1 \pmod{4} \\
     -1 & \text{if } i \equiv 2 \pmod{4} \\
     10 & \text{if } i \equiv 3 \pmod{4} \\
     1 & \text{if } i \equiv 0 \pmod{4}
     \end{cases} \pmod{101}
     \]
   - Therefore, the condition for \( m \) to be divisible by 101 is:
     \[
     m \equiv -10A - B + 10C + D \equiv 10(C - A) + (D - B) \equiv 0 \pmod{101}
     \]

5. **Counting valid configurations:**
   - We need \( C = A \) and \( D = B \) for the simplest case. Let \( Q(k) \) be the number of ways to choose \( a_i \) such that \( C = A \) and \( D = B \).
   - To have \( A = C \), we need an equal number of 2's in both sums. Since \( a_1 = 2 \), we discard it from \( A \) and balance the remaining sums. This can be done in:
     \[
     \sum_{r=0}^{k-1} \binom{k}{r+1} \binom{k-1}{r} = \binom{2k-1}{k}
     \]
     ways using Vandermonde's identity.
   - Similarly, the number of ways to pick \( a_i \) such that \( B = D \) is also \( \binom{2k-1}{k} \).

6. **Combining the results:**
   - Therefore, \( Q(k) = \binom{2k-1}{k}^2 \), and since \( P(k) \geq Q(k) \), we have:
     \[
     P(k) \geq \binom{2k-1}{k}^2
     \]

7. **Determining when equality holds:**
   - We need to check when \( 10(C - A) + (D - B) \equiv 0 \pmod{101} \) forces \( C = A \) and \( D = B \).
   - For \( k \leq 9 \), the range of \( 10(C - A) + (D - B) \) is small enough that it must be zero, implying \( C = A \) and \( D = B \).
   - For \( k > 9 \), there are configurations where \( 10(C - A) + (D - B) \equiv 0 \pmod{101} \) without \( C = A \) and \( D = B \).

The final answer is \( \boxed{ k \leq 9 } \)