Find all positive integers $n$ such that the number
$$n^6 + 5n^3 + 4n + 116$$
is the product of two or more consecutive numbers.
To find all positive integers \( n \) such that the number \( n^6 + 5n^3 + 4n + 116 \) is the product of two or more consecutive numbers, we will consider different cases based on the number of consecutive numbers.

1. **Product of exactly two consecutive numbers:**

   Let \( n^6 + 5n^3 + 4n + 116 = k(k+1) \).

   This can be rewritten as:
   \[
   n^6 + 5n^3 + 4n + 116 = k^2 + k
   \]

   Multiplying both sides by 4 and adding 1, we get:
   \[
   4(n^6 + 5n^3 + 4n + 116) + 1 = 4k^2 + 4k + 1
   \]

   This simplifies to:
   \[
   4n^6 + 20n^3 + 16n + 465 = (2k+1)^2
   \]

   Let \( 2k + 1 = m \), where \( m \) is an odd integer. Then:
   \[
   4n^6 + 20n^3 + 16n + 465 = m^2
   \]

   Since \( m \) is odd, let \( m = 2n^3 + 2a + 1 \) for some integer \( a \ge 0 \). Substituting \( m \) into the equation, we get:
   \[
   (2n^3 + 2a + 1)^2 = 4n^6 + 20n^3 + 16n + 465
   \]

   Expanding the left-hand side:
   \[
   4n^6 + 8n^3a + 4a^2 + 4n^3 + 4a + 1 = 4n^6 + 20n^3 + 16n + 465
   \]

   Simplifying, we get:
   \[
   4a^2 + 8n^3a + 4a + 1 = 16n + 465
   \]

   Rearranging terms:
   \[
   4a^2 + 8n^3a + 4a - 16n - 464 = 0
   \]

   Dividing by 4:
   \[
   a^2 + 2n^3a + a - 4n - 116 = 0
   \]

   For \( a \ge 11 \), the left-hand side is larger than the right-hand side. For \( 0 \le a \le 10 \), we check each value of \( a \):

   - For \( a = 4 \):
     \[
     4^2 + 2n^3 \cdot 4 + 4 - 4n - 116 = 0
     \]
     \[
     16 + 8n^3 + 4 - 4n - 116 = 0
     \]
     \[
     8n^3 - 4n - 96 = 0
     \]
     \[
     2n^3 - n - 12 = 0
     \]

     Testing \( n = 3 \):
     \[
     2(3)^3 - 3 - 12 = 0
     \]
     \[
     54 - 3 - 12 = 0
     \]
     \[
     39 \neq 0
     \]

     Thus, \( n = 3 \) is not a solution.

2. **Product of five or more consecutive numbers:**

   The product of at least five consecutive numbers is a multiple of 5, whereas \( n^6 + 5n^3 + 4n + 116 \) is never a multiple of 5. Therefore, there are no solutions in this case.

3. **Product of exactly three consecutive numbers:**

   Let \( n^6 + 5n^3 + 4n + 116 = (k-1)k(k+1) \).

   Substituting \( k = n^2 + b \) for some integer \( b \), we get:
   \[
   n^6 + 5n^3 + 4n + 116 = (n^2 + b - 1)(n^2 + b)(n^2 + b + 1)
   \]

   Expanding the right-hand side:
   \[
   (n^2 + b - 1)(n^2 + b)(n^2 + b + 1) = (n^2 + b)^3 - (n^2 + b)
   \]

   Simplifying:
   \[
   (n^2 + b)^3 - (n^2 + b) = n^6 + 3n^4b + 3n^2b^2 + b^3 - n^2 - b
   \]

   Equating both sides:
   \[
   n^6 + 3n^4b + 3n^2b^2 + b^3 - n^2 - b = n^6 + 5n^3 + 4n + 116
   \]

   Simplifying:
   \[
   3n^4b + 3n^2b^2 + b^3 - n^2 - b = 5n^3 + 4n + 116
   \]

   For \( b \ge 5 \), the left-hand side is larger than the right-hand side. For \( b \le 0 \), the left-hand side is non-positive and the right-hand side is positive. For \( 1 \le b \le 4 \), there is no solution.

4. **Product of exactly four consecutive numbers:**

   Let \( n^6 + 5n^3 + 4n + 116 = k(k+1)(k+2)(k+3) \).

   This can be rewritten as:
   \[
   n^6 + 5n^3 + 4n + 116 = (k^2 + 3k + 1)^2 - 1
   \]

   Adding 1 to both sides:
   \[
   n^6 + 5n^3 + 4n + 117 = (k^2 + 3k + 1)^2
   \]

   Let \( k^2 + 3k + 1 = n^3 + c \) for some integer \( c \ge 1 \). Then:
   \[
   (n^3 + c)^2 = n^6 + 5n^3 + 4n + 117
   \]

   Expanding the left-hand side:
   \[
   n^6 + 2n^3c + c^2 = n^6 + 5n^3 + 4n + 117
   \]

   Simplifying:
   \[
   2n^3c + c^2 = 5n^3 + 4n + 117
   \]

   Rearranging terms:
   \[
   c^2 + (2c - 5)n^3 = 4n + 117
   \]

   For \( c \ge 11 \), the left-hand side is larger than the right-hand side. For \( 1 \le c \le 10 \), there are no integer solutions.

The final answer is \( \boxed{ n = 3 } \).