Find all sets comprising of 4 natural numbers such that the product of any 3 numbers in the set leaves a remainder of 1 when divided by the remaining number.
1. **Given Condition**: We need to find all sets of 4 natural numbers \(\{a, b, c, d\}\) such that the product of any 3 numbers in the set leaves a remainder of 1 when divided by the remaining number. Mathematically, this can be expressed as:
   \[
   \frac{abc}{d} \equiv 1 \pmod{d}, \quad \frac{abd}{c} \equiv 1 \pmod{c}, \quad \frac{acd}{b} \equiv 1 \pmod{b}, \quad \frac{bcd}{a} \equiv 1 \pmod{a}
   \]

2. **Rewriting the Conditions**: The above conditions can be rewritten as:
   \[
   abc \equiv d \pmod{d}, \quad abd \equiv c \pmod{c}, \quad acd \equiv b \pmod{b}, \quad bcd \equiv a \pmod{a}
   \]
   This implies:
   \[
   abc \equiv 1 \pmod{d}, \quad abd \equiv 1 \pmod{c}, \quad acd \equiv 1 \pmod{b}, \quad bcd \equiv 1 \pmod{a}
   \]

3. **Summing the Conditions**: Summing all the conditions, we get:
   \[
   abc + abd + acd + bcd \equiv 4 \pmod{abcd}
   \]
   This implies:
   \[
   abcd \mid (abc + abd + acd + bcd - 4)
   \]

4. **Function Analysis**: Define the function:
   \[
   f(a, b, c, d) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{abcd}
   \]
   We need \(f(a, b, c, d) \geq 1\).

5. **Minimum Values**: To satisfy \(f(a, b, c, d) \geq 1\), we need to find the minimum values of \(a, b, c, d\). Since \(f(a, b, c, d)\) is decreasing in each variable, we start with the smallest natural numbers:
   \[
   \min(a, b, c, d) = 1, 2, 3, 4
   \]

6. **Checking Possible Values**: Plugging in \(a = 1, 2, 3, 4\) and finding possible values for the remaining numbers:
   - For \(a = 1\):
     \[
     \frac{1}{1} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{bcd} \geq 1
     \]
     Simplifying, we get:
     \[
     1 + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{bcd} \geq 1
     \]
     This is always true since \(1 + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{bcd} > 1\).

   - For \(a = 2\):
     \[
     \frac{1}{2} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{2bcd} \geq 1
     \]
     Simplifying, we get:
     \[
     \frac{1}{2} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{2bcd} \geq 1
     \]
     This is true for \(b, c, d \geq 2\).

   - For \(a = 3\):
     \[
     \frac{1}{3} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{3bcd} \geq 1
     \]
     Simplifying, we get:
     \[
     \frac{1}{3} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{3bcd} \geq 1
     \]
     This is true for \(b, c, d \geq 3\).

   - For \(a = 4\):
     \[
     \frac{1}{4} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{4bcd} \geq 1
     \]
     Simplifying, we get:
     \[
     \frac{1}{4} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{4bcd} \geq 1
     \]
     This is true for \(b, c, d \geq 4\).

7. **Conclusion**: The sets of 4 natural numbers that satisfy the given condition are \(\{1, 2, 3, 4\}\).

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