For how many positive integers $x$ less than $4032$ is $x^2-20$ divisible by $16$ and $x^2-16$ divisible by $20$?

[i] Proposed by Tristan Shin [/i]

To solve the problem, we need to find the number of positive integers \( x \) less than \( 4032 \) such that \( x^2 - 20 \) is divisible by \( 16 \) and \( x^2 - 16 \) is divisible by \( 20 \).

1. **Set up the congruences:**
   \[
   x^2 - 20 \equiv 0 \pmod{16} \implies x^2 \equiv 20 \pmod{16} \implies x^2 \equiv 4 \pmod{16}
   \]
   \[
   x^2 - 16 \equiv 0 \pmod{20} \implies x^2 \equiv 16 \pmod{20}
   \]

2. **Solve the system of congruences:**
   Let \( y = x^2 \). We need to solve:
   \[
   y \equiv 4 \pmod{16}
   \]
   \[
   y \equiv 16 \pmod{20}
   \]

3. **Use the Chinese Remainder Theorem (CRT):**
   We need to find \( y \) such that:
   \[
   y \equiv 4 \pmod{16}
   \]
   \[
   y \equiv 16 \pmod{20}
   \]

   Since \( y \equiv 16 \pmod{20} \), we can write:
   \[
   y = 20k + 16 \text{ for some integer } k
   \]

   Substitute into the first congruence:
   \[
   20k + 16 \equiv 4 \pmod{16}
   \]
   \[
   20k + 16 \equiv 4 \pmod{16} \implies 4k + 0 \equiv 4 \pmod{16} \implies 4k \equiv 4 \pmod{16}
   \]
   \[
   k \equiv 1 \pmod{4}
   \]

   Therefore, \( k = 4m + 1 \) for some integer \( m \). Substitute back:
   \[
   y = 20(4m + 1) + 16 = 80m + 36
   \]
   \[
   y \equiv 36 \pmod{80}
   \]

4. **Solve for \( x \):**
   We need \( x^2 \equiv 36 \pmod{80} \). This implies:
   \[
   x \equiv \pm 6 \pmod{80}
   \]
   \[
   x \equiv 6 \pmod{80} \quad \text{or} \quad x \equiv 74 \pmod{80}
   \]

5. **Count the solutions for \( x < 4032 \):**
   For \( x \equiv 6 \pmod{80} \):
   \[
   x = 80k + 6 \quad \text{for} \quad k = 0, 1, 2, \ldots, 50
   \]
   The largest \( k \) such that \( 80k + 6 < 4032 \) is:
   \[
   80k + 6 < 4032 \implies 80k < 4026 \implies k < 50.325 \implies k \leq 50
   \]
   Thus, there are \( 51 \) solutions for \( x \equiv 6 \pmod{80} \).

   For \( x \equiv 74 \pmod{80} \):
   \[
   x = 80k + 74 \quad \text{for} \quad k = 0, 1, 2, \ldots, 49
   \]
   The largest \( k \) such that \( 80k + 74 < 4032 \) is:
   \[
   80k + 74 < 4032 \implies 80k < 3958 \implies k < 49.475 \implies k \leq 49
   \]
   Thus, there are \( 50 \) solutions for \( x \equiv 74 \pmod{80} \).

6. **Total number of solutions:**
   \[
   51 + 50 = 101
   \]

The final answer is \(\boxed{101}\).