$$Problem 1$$ ;Find all composite numbers $n$ with the following property: For every proper divisor $d$ of $n$ (i.e. $1 < d < n$), it holds that $n-12 \geq d \geq n-20$.
To solve this problem, we need to find all composite numbers \( n \) such that for every proper divisor \( d \) of \( n \) (i.e., \( 1 < d < n \)), the inequality \( n-12 \geq d \geq n-20 \) holds.

1. **Understanding the inequality**:
   - For every proper divisor \( d \) of \( n \), we have:
     \[
     n-12 \geq d \geq n-20
     \]
   - This implies that all proper divisors \( d \) of \( n \) must lie within the interval \([n-20, n-12]\).

2. **Analyzing the interval**:
   - The interval \([n-20, n-12]\) has a length of \( 8 \) (i.e., \( 20 - 12 = 8 \)).
   - Therefore, all proper divisors of \( n \) must be within an 8-unit range.

3. **Considering the smallest possible \( n \)**:
   - Since \( n \) is composite, it must have at least one proper divisor other than 1 and itself.
   - Let's start with the smallest composite number, \( n = 4 \):
     - Proper divisors of 4: \( 2 \)
     - Check the inequality: \( 4-12 \leq 2 \leq 4-20 \) which simplifies to \( -8 \leq 2 \leq -16 \). This is not possible.
   - Next, consider \( n = 6 \):
     - Proper divisors of 6: \( 2, 3 \)
     - Check the inequality: \( 6-12 \leq 2 \leq 6-20 \) and \( 6-12 \leq 3 \leq 6-20 \). This simplifies to \( -6 \leq 2 \leq -14 \) and \( -6 \leq 3 \leq -14 \). This is not possible.
   - Continue this process for larger composite numbers.

4. **Finding a valid \( n \)**:
   - Let's try \( n = 24 \):
     - Proper divisors of 24: \( 2, 3, 4, 6, 8, 12 \)
     - Check the inequality for each divisor:
       \[
       24-12 \geq 2 \geq 24-20 \implies 12 \geq 2 \geq 4 \quad \text{(not possible)}
       \]
       \[
       24-12 \geq 3 \geq 24-20 \implies 12 \geq 3 \geq 4 \quad \text{(not possible)}
       \]
       \[
       24-12 \geq 4 \geq 24-20 \implies 12 \geq 4 \geq 4 \quad \text{(possible)}
       \]
       \[
       24-12 \geq 6 \geq 24-20 \implies 12 \geq 6 \geq 4 \quad \text{(possible)}
       \]
       \[
       24-12 \geq 8 \geq 24-20 \implies 12 \geq 8 \geq 4 \quad \text{(possible)}
       \]
       \[
       24-12 \geq 12 \geq 24-20 \implies 12 \geq 12 \geq 4 \quad \text{(possible)}
       \]
     - All proper divisors of 24 satisfy the inequality.

5. **Conclusion**:
   - The only composite number \( n \) that satisfies the given condition is \( n = 24 \).

The final answer is \( \boxed{24} \)