In a group of $n > 20$ people, there are some (at least one, and possibly all) pairs of people that know each other. Knowing is symmetric; if Alice knows Blaine, then Blaine also knows Alice. For some values of $n$ and $k,$ this group has a peculiar property: If any $20$ people are removed from the group, the number of pairs of people that know each other is
at most $\frac{n-k}{n}$ times that of the original group of people.
(a) If $k = 41,$ for what positive integers $n$ could such a group exist?
(b) If $k = 39,$ for what positive integers $n$ could such a group exist?
1. **Define the problem and notation:**
   - Let a **connection** be a pair of people who know each other.
   - Let there be \( C \) connections among all \(\binom{n}{2}\) pairs of people.
   - Let the \(\binom{n}{20}\) \((n-20)\)-person subsets of the \( n \) people be named subset \( 1 \), subset \( 2 \), \(\cdots\), subset \(\binom{n}{20}\).
   - For \( 1 \le x \le \binom{n}{20} \), let \( F(x) \) denote the number of connections among the people in subset \( x \).

2. **Sum of connections in all subsets:**
   - We know that 
     \[
     F(1) + F(2) + \cdots + F\left(\binom{n}{20}\right) = C \cdot \binom{n-2}{20}
     \]
     because if we take one individual connection (between person \( A \) and person \( B \)), the number of times it is counted in the above sum is the number of ways to include \( A \) and \( B \) in the \( n-20 \) person subset, or the number of ways to NOT include \( A \) or \( B \) among the 20 people being removed.

3. **Inequality for each subset:**
   - If every term in \(\{F(1), \cdots, F\left(\binom{n}{20}\right)\}\) is \(\le C \cdot \frac{n-k}{n}\), then we know 
     \[
     C \cdot \binom{n-2}{20} \le C \cdot \frac{n-k}{n} \cdot \binom{n}{20}
     \]
     Simplifying, we get:
     \[
     \binom{n-2}{20} \le \frac{n-k}{n} \cdot \binom{n}{20}
     \]
     \[
     \frac{(n-20)(n-21)}{(n)(n-1)} \le \frac{n-k}{n}
     \]
     \[
     (n-20)(n-21) \le (n-1)(n-k)
     \]

4. **Part (a): \( k = 41 \):**
   - Substituting \( k = 41 \) into the inequality:
     \[
     (n-20)(n-21) \le (n-1)(n-41)
     \]
     \[
     n^2 - 41n + 420 \le n^2 - 42n + 41
     \]
     \[
     n \le -379
     \]
     Since \( n > 20 \), this case is impossible.

5. **Part (b): \( k = 39 \):**
   - Substituting \( k = 39 \) into the inequality:
     \[
     (n-20)(n-21) \le (n-1)(n-39)
     \]
     \[
     n^2 - 41n + 420 \le n^2 - 40n + 39
     \]
     \[
     n \ge 381
     \]
   - Now let's check that all \( n \ge 381 \) work. Consider the case in which every possible pair of people know each other. Then there are \(\binom{n}{2}\) connections among all \( n \) people, and in a group of \( n-20 \) people, there are exactly \(\binom{n-20}{2}\) connections. If \( n \ge 381 \), then:
     \[
     \frac{(n-20)(n-21)}{(n)(n-1)} \le \frac{n-39}{n}
     \]
     \[
     \binom{n-20}{2} \div \binom{n}{2} \le \frac{n-39}{n}
     \]
     \[
     \binom{n-20}{2} \le \binom{n}{2} \cdot \frac{n-39}{n}
     \]
   - Thus, the number of connections in every group of \( n-20 \) people is at most \(\frac{n-39}{n}\) times the number of connections among all \( n \) people, and every value \( \boxed{n \ge 381} \) is possible.