Given the set $\{n, n+1, n+2, n+3, n+4, n+5\}$, we are tasked with determining whether it can be partitioned into two subsets such that the product of the numbers in each subset is equal.

To solve this problem, let's denote the two subsets as $A$ and $B$. The condition requires:
$$\prod_{a \in A} a = \prod_{b \in B} b.$$

Since there are 6 elements in total, each subset must contain 3 elements. Without loss of generality, assume:
$$A = \{n, n+1, n+2\}, \quad B = \{n+3, n+4, n+5\}.$$

Then the condition becomes:
$$n(n+1)(n+2) = (n+3)(n+4)(n+5).$$

We will analyze whether this equality holds for positive integers.

### Expanding the Products

1. Expand the product for $A$:
$$n(n+1)(n+2) = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n.$$

2. Expand the product for $B$:
$$(n+3)(n+4)(n+5) = (n+3)((n+4)(n+5)).$$
First, calculate:
$$(n+4)(n+5) = n^2 + 9n + 20.$$
Then multiply by $(n+3)$:
$$(n+3)(n^2 + 9n + 20) = n^3 + 9n^2 + 20n + 3n^2 + 27n + 60 = n^3 + 12n^2 + 47n + 60.$$

### Comparing the Expressions

We equate the two expressions:
$$n^3 + 3n^2 + 2n = n^3 + 12n^2 + 47n + 60.$$

Simplifying, we get:
$$0 = 9n^2 + 45n + 60.$$

Divide throughout by 3:
$$0 = 3n^2 + 15n + 20.$$

Calculating the discriminant of the quadratic:
$$\Delta = 15^2 - 4 \cdot 3 \cdot 20 = 225 - 240 = -15.$$

The discriminant $\Delta$ is negative, indicating there are no real roots, thus no integer solutions exist.

### Conclusion

There are no positive integer values for $n$ that allow the set $\{n, n+1, n+2, n+3, n+4, n+5\}$ to be partitioned into two subsets with equal products of their elements. Therefore, the solution is:
$$\boxed{\text{No solution}}$$