A sum of $335$ pairwise distinct positive integers equals $100000$.
a) What is the least number of uneven integers in that sum?
b) What is the greatest number of uneven integers in that sum?
Given the problem, we need to find the least and greatest number of odd integers in a sum of 335 pairwise distinct positive integers that equals 100,000.

1. **Define Variables:**
   Let \( n \) be the number of odd integers and \( 335 - n \) be the number of even integers.

2. **Sum Condition:**
   The sum of these integers must be 100,000. Since the sum is even, the number of odd integers \( n \) must be even (because the sum of an odd number of odd integers is odd, and the sum of an even number of odd integers is even).

3. **Sum of Odd and Even Integers:**
   The sum of the first \( n \) odd integers is:
   \[
   \sum_{k=1}^n (2k-1) = n^2
   \]
   The sum of the first \( 335 - n \) even integers is:
   \[
   \sum_{k=1}^{335-n} 2k = 2 \sum_{k=1}^{335-n} k = 2 \cdot \frac{(335-n)(336-n)}{2} = (335-n)(336-n)
   \]

4. **Total Sum Equation:**
   Combining these, we get:
   \[
   n^2 + (335-n)(336-n) = 100,000
   \]

5. **Simplify the Equation:**
   \[
   n^2 + 112560 - 671n + n^2 = 100,000
   \]
   \[
   2n^2 - 671n + 12560 \leq 0
   \]

6. **Solve the Quadratic Inequality:**
   To find the range of \( n \), solve the quadratic equation:
   \[
   2n^2 - 671n + 12560 = 0
   \]
   Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   n = \frac{671 \pm \sqrt{671^2 - 4 \cdot 2 \cdot 12560}}{2 \cdot 2}
   \]
   \[
   n = \frac{671 \pm \sqrt{450241 - 100480}}{4}
   \]
   \[
   n = \frac{671 \pm \sqrt{349761}}{4}
   \]
   \[
   n = \frac{671 \pm 591.43}{4}
   \]
   \[
   n_1 = \frac{671 + 591.43}{4} \approx 315.11
   \]
   \[
   n_2 = \frac{671 - 591.43}{4} \approx 19.89
   \]

7. **Determine the Bounds:**
   Since \( n \) must be even:
   \[
   20 \leq n \leq 314
   \]

8. **Verification:**
   - For \( n = 20 \):
     \[
     \sum_{k=1}^{20} (2k-1) = 20^2 = 400
     \]
     \[
     \sum_{k=1}^{315} 2k = 315 \cdot 316 = 99440
     \]
     \[
     400 + 99440 = 99840
     \]
   - For \( n = 314 \):
     \[
     \sum_{k=1}^{314} (2k-1) = 314^2 = 98596
     \]
     \[
     \sum_{k=1}^{21} 2k = 21 \cdot 22 = 462
     \]
     \[
     98596 + 462 = 99058
     \]

Thus, the least number of odd integers is 20, and the greatest number of odd integers is 314.

The final answer is \( \boxed{ n \geq 20 \text{ and } n \leq 314 } \)