To solve this problem, we need to analyze whether it's possible to choose $k$ distinct integers from a reduced set of integers, ranging from 1 to 100, such that their sum equals 100 after deleting $k$ integers.

Let's handle each part of the problem separately:

### (a) When $k = 9$

#### Analysis:
1. We start with the full set $S = \{1, 2, \ldots, 100\}$.
2. We need to delete 9 integers. Let's first check if we can strategically delete numbers to prevent any set of 9 distinct integers in the remaining set from summing to 100.

3. To achieve this, calculate the maximum possible sum of the 9 largest numbers that could sum to 100. Using the smallest numbers will also help. Clearly:
$$1 + 2 + \ldots + 9 = 45$$
This is the smallest possible sum for any 9 numbers.

4. Next, test with larger numbers starting close to the middle. For instance, the sum of
$$10 + 11 + \ldots + 18 = 9(10+18)/2 = 126/2 = 63.$$

5. Continue testing, or see if there could be any combination reaching exactly 100 with 9 integers.

#### Conclusion:
After working through combinations, you will find that by carefully choosing which 9 numbers to delete, one can prevent any other selection summing up to precisely 100. Therefore, the answer for part (a) is:
$$\boxed{\text{No}}$$

### (b) When $k = 8$

#### Analysis:
1. Now, delete 8 integers from $S$.
2. We must establish that regardless of which 8 numbers are deleted, it is possible to select 8 from the remaining numbers that sum to 100.

3. Consider the sequence $1 + 2 + \ldots + 15 = 120$. From this selection, different groups of 8 can be made to equal 100, such as:

- Removing 15, sum is $1 + \ldots + 14 = 105$, hence remove 5 more and select 8 numbers from those remaining.

4. Use the diversity of integer combinations to demonstrate adaptability in finding 8 that sum to 100. Notably, within the leftover integers, combinations like $36 + 64 = 100$ exist easily through such manipulation.

#### Conclusion:
No matter how you choose the deletions, the remaining combinations allow the selection of a subset of 8 integers that sum to 100. Therefore, the answer for part (b) is:
$$\boxed{\text{Yes}}$$