It is given set $A=\{1,2,3,...,2n-1\}$. From set $A$, at least $n-1$ numbers are expelled such that:
$a)$ if number $a \in A$ is expelled, and if $2a \in A$ then $2a$ must be expelled
$b)$ if $a,b \in A$ are expelled, and $a+b \in A$ then $a+b$ must be also expelled
Which numbers must be expelled such that sum of numbers remaining in set stays minimal
To solve this problem, we need to expel numbers from the set \( A = \{1, 2, 3, \ldots, 2n-1\} \) such that the sum of the remaining numbers is minimized while adhering to the given conditions. Let's break down the steps:

1. **Identify the conditions for expulsion:**
   - If a number \( a \in A \) is expelled, and if \( 2a \in A \), then \( 2a \) must also be expelled.
   - If \( a, b \in A \) are expelled, and \( a + b \in A \), then \( a + b \) must also be expelled.

2. **Determine the numbers to expel:**
   - We need to expel at least \( n-1 \) numbers.
   - To minimize the sum of the remaining numbers, we should aim to expel the largest numbers possible, as larger numbers contribute more to the sum.

3. **Expel numbers from \( 2 \) to \( 2n-1 \):**
   - If we expel all numbers from \( 2 \) to \( 2n-1 \), we are left with only the number \( 1 \).
   - This satisfies the condition of expelling at least \( n-1 \) numbers because there are \( 2n-2 \) numbers in the range \( 2 \) to \( 2n-1 \), which is more than \( n-1 \).

4. **Verify the conditions:**
   - If \( a = 1 \) is not expelled, then \( 2a = 2 \) must be expelled, which is already included in our expulsion list.
   - If any two numbers \( a \) and \( b \) are expelled, their sum \( a + b \) will also be expelled if it lies within the range \( 2 \) to \( 2n-1 \), which is true for our expulsion list.

Thus, expelling all numbers from \( 2 \) to \( 2n-1 \) ensures that the sum of the remaining numbers is minimized while satisfying the given conditions.

The final answer is \( \boxed{ \{2, 3, \ldots, 2n-1\} } \)