At least $ n - 1$ numbers are removed from the set $\{1, 2, \ldots, 2n - 1\}$ according to the following rules:
(i) If $ a$ is removed, so is $ 2a$;
(ii) If $ a$ and $ b$ are removed, so is $ a \plus{} b$.
Find the way of removing numbers such that the sum of the remaining numbers is maximum possible.
To solve this problem, we need to find a way to remove at least \( n-1 \) numbers from the set \( A = \{1, 2, \ldots, 2n-1\} \) such that the sum of the remaining numbers is maximized. We will use the given rules:
1. If \( a \) is removed, then \( 2a \) must also be removed.
2. If \( a \) and \( b \) are removed, then \( a + b \) must also be removed.

Let's analyze the problem step-by-step:

1. **Removing the smallest number:**
   - If we remove 1, then we must also remove 2 (since \( 2 \times 1 = 2 \)), 3 (since \( 1 + 2 = 3 \)), and so on. This will lead to removing all numbers in the set \( A \). Therefore, removing 1 is not optimal.

2. **Removing even numbers:**
   - Consider removing the even numbers \( 2, 4, 6, \ldots, 2(n-1) \). This removes \( n-1 \) numbers. According to the rules, removing these numbers does not force us to remove any additional numbers because:
     - \( 2a \) for any \( a \) in the set of even numbers is already included in the set of even numbers.
     - The sum of any two even numbers is also an even number, which is already included in the set of even numbers.

3. **Sum of remaining numbers:**
   - The sum of the set \( A \) is:
     \[
     \sum_{k=1}^{2n-1} k = \frac{(2n-1) \cdot 2n}{2} = n(2n-1)
     \]
   - The sum of the removed even numbers is:
     \[
     \sum_{k=1}^{n-1} 2k = 2 \sum_{k=1}^{n-1} k = 2 \cdot \frac{(n-1)n}{2} = n(n-1)
     \]
   - Therefore, the sum of the remaining numbers is:
     \[
     n(2n-1) - n(n-1) = n(2n-1) - n^2 + n = n^2
     \]

4. **Verification of optimality:**
   - By Lemma 1, removing 1 leads to removing all numbers, which is not optimal.
   - By Lemma 2, removing the even numbers \( 2, 4, \ldots, 2(n-1) \) leaves us with a sum of \( n^2 \).
   - By Lemma 3, any other combination of removed numbers will result in a sum of the remaining numbers that is less than \( n^2 \).

Thus, the optimal way to remove numbers is to remove the even numbers \( 2, 4, 6, \ldots, 2(n-1) \), which leaves the sum of the remaining numbers as \( n^2 \).

The final answer is \( \boxed{ n^2 } \)