In $n$ transparent boxes there are red balls and blue balls. One needs to choose $50$ boxes such that, together, they contain at least half of the red balls and at least half of the blue balls. Is such a choice possible irrespective on the number of balls and on the way they are distributed in the boxes, if:
a) $n = 100$
b) $n = 99$?
To solve the problem, we need to determine if it is possible to choose 50 boxes out of \( n \) boxes such that they contain at least half of the red balls and at least half of the blue balls, regardless of the number of balls and their distribution.

### Part (a): \( n = 100 \)

1. **Restate the problem in terms of a known result:**
   The problem is similar to a known result in combinatorics, which states that in \( 2n \) boxes containing two types of items, it is always possible to choose \( n+1 \) boxes such that they contain at least half of each type of item.

2. **Apply the known result:**
   For \( n = 100 \), we have \( 2n = 200 \) boxes. According to the known result, we can choose \( n+1 = 101 \) boxes to ensure that they contain at least half of each type of item.

3. **Adjust the number of boxes:**
   Since we only have 100 boxes, we need to check if choosing 50 boxes (half of 100) can still satisfy the condition. The known result does not directly apply to choosing half the boxes, but we can use a similar argument.

4. **Consider the worst-case distribution:**
   Suppose the red balls and blue balls are distributed in such a way that each box contains an equal number of red and blue balls. In this case, choosing any 50 boxes will contain exactly half of the red balls and half of the blue balls.

5. **Generalize the argument:**
   Even if the distribution is not equal, by the pigeonhole principle, choosing 50 boxes out of 100 will still capture at least half of the total number of red balls and blue balls, because we are choosing half of the total number of boxes.

Therefore, it is possible to choose 50 boxes out of 100 such that they contain at least half of the red balls and at least half of the blue balls.

### Part (b): \( n = 99 \)

1. **Restate the problem in terms of a known result:**
   For \( n = 99 \), we have \( 2n = 198 \) boxes. According to the known result, we can choose \( n+1 = 100 \) boxes to ensure that they contain at least half of each type of item.

2. **Adjust the number of boxes:**
   Since we only have 99 boxes, we need to check if choosing 50 boxes (slightly more than half of 99) can still satisfy the condition. The known result does not directly apply to choosing half the boxes, but we can use a similar argument.

3. **Consider the worst-case distribution:**
   Suppose the red balls and blue balls are distributed in such a way that each box contains an equal number of red and blue balls. In this case, choosing any 50 boxes will contain slightly more than half of the red balls and blue balls.

4. **Generalize the argument:**
   Even if the distribution is not equal, by the pigeonhole principle, choosing 50 boxes out of 99 will still capture at least half of the total number of red balls and blue balls, because we are choosing slightly more than half of the total number of boxes.

Therefore, it is possible to choose 50 boxes out of 99 such that they contain at least half of the red balls and at least half of the blue balls.

The final answer is True for both \( \boxed{ n = 100 } \) and \( n = 99 \).