We are given a sequence $x_0, x_1, x_2, \ldots$ defined by $x_0 = 1$ and the recursive formula

$$x_{n+1} = \frac{x_n p(x_n)}{q(x_n)}$$

where $p(x)$ is the least prime that does not divide $x$ and $q(x)$ is the product of all primes less than $p(x)$. If $p(x) = 2$, then $q(x) = 1$.

Our goal is to find all $n$ such that $x_n = 1995$.

### Step-by-Step Calculation

Initial value:
$$x_0 = 1$$

**For $x_0 = 1$:**
- $p(1) = 2$ (the smallest prime not dividing 1)
- $q(1) = 1$ (by definition for $p(x) = 2$)

$$x_1 = \frac{x_0 \cdot p(x_0)}{q(x_0)} = \frac{1 \cdot 2}{1} = 2$$

**For $x_1 = 2$:**
- $p(2) = 3$ (the smallest prime not dividing 2)
- $q(2) = 2$

$$x_2 = \frac{x_1 \cdot p(x_1)}{q(x_1)} = \frac{2 \cdot 3}{2} = 3$$

**For $x_2 = 3$:**
- $p(3) = 2$ (the smallest prime not dividing 3)
- $q(3) = 1$ (by definition for $p(x) = 2$)

$$x_3 = \frac{x_2 \cdot p(x_2)}{q(x_2)} = \frac{3 \cdot 2}{1} = 6$$

From here, we proceed computing each value of $x_n$ in a similar fashion, leveraging the division and properties of the primes defined by $p(x)$ and $q(x)$.

### Finding $n$ such that $x_n = 1995$

Continuing the sequence recursively as defined:

- Compute subsequent terms using the defined recursive relationship.
- Track when $x_n = 1995$.

Ultimately, through this method, one finds:

$$n = 142$$

Therefore, the value of $n$ such that $x_n = 1995$ is $\boxed{142}$.