Find all positive integers $n$ such that $n!$ can be written as the product of two Fibonacci numbers.
1. **Assume the Fibonacci sequence**: 
   The Fibonacci sequence is defined as \( F_1 = 1 \), \( F_2 = 1 \), and \( F_{n} = F_{n-1} + F_{n-2} \) for \( n \geq 3 \).

2. **Estimate Fibonacci numbers**:
   We can use the approximation \( F_k \geq (1.6)^{k-1} \) for \( k \geq 2 \). This is derived from the fact that the ratio of consecutive Fibonacci numbers approaches the golden ratio \( \phi \approx 1.618 \).

3. **Divisibility properties**:
   We use the property that \( \frac{F_{3k}}{F_k} = 5F_k^2 + 3(-1)^k \). This helps in understanding the divisibility of Fibonacci numbers by 3.

4. **Lemma on divisibility by 3**:
   If \( v_3(F_k) \geq t \), then \( 4 \cdot 3^{t-1} \) divides \( k \). This lemma is proved by induction on \( t \).

5. **Divisibility of factorials**:
   We know that \( v_3(n!) \geq \frac{n}{3} \). This is derived from Legendre's formula for the highest power of a prime \( p \) dividing \( n! \).

6. **Assume \( n! = F_k \cdot F_m \)**:
   Let \( x = v_3(F_k) \geq 0 \) and \( y = v_3(F_m) \geq 2 \). Then \( x + y = v_3(n!) \geq \frac{n}{3} \).

7. **Bounds on \( k \) and \( m \)**:
   Using the lemma, we get \( k \geq 1 + 3^x \) and \( m \geq 4 \cdot 3^{y-1} \geq 3^y + 3 \).

8. **Inequality involving factorials and Fibonacci numbers**:
   \[
   2^{n(n-1)} \geq n^{n-1} \geq n! = F_k \cdot F_m \geq 6^{3^x + 3^y + 2} \geq 6^{2 \cdot 3^{\frac{x+y}{2} + 2}} \geq 2^{3^{\frac{n}{6} + 2}}
   \]
   This implies \( n(n-1) \geq 3^{\frac{n}{6} + 2} \).

9. **Solving the inequality**:
   From the inequality, we deduce that \( n \geq 20 \). However, since \( 20! \geq F_n \), we need to check smaller values of \( n \).

10. **Checking smaller values**:
    For \( n \leq 20 \), we find that the only values of \( n \) such that \( n! \) can be written as the product of two Fibonacci numbers are \( n = 1, 2, 3, 4, 6 \).

The final answer is \( \boxed{ n = 1, 2, 3, 4, 6 } \).