Given a positive integer $n$, we are tasked with determining if there exists a positive integer $n$ for which the denominator $g(n)$ of the rational representation of the sum

$$1 + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!} = \frac{f(n)}{g(n)}$$

satisfies $g(n) > n^{0.999n}$.

### Step-by-Step Analysis

1. Expression for the Sum:
The given series represents the partial sum of the exponential function's series expansion up to the $n$-th term. The series is

$$S_n = 1 + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!}.$$

This sum can be written as a single fraction:

$$S_n = \frac{f(n)}{g(n)}$$

where both $f(n)$ and $g(n)$ are integers and $\gcd(f(n), g(n)) = 1$.

2. Approximation and Properties:
The series $S_n$ approaches the value of $e$ (Euler's number) as $n$ increases. This is evident because

$$S_n = e - \left( \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots \right).$$

While evaluating the denominator $g(n)$, note that each $\frac{1}{k!}$ can be expressed with $k!$ as a common denominator. Consequently,

$$S_n = \frac{\left(\prod_{k=1}^{n} k! \right)}{n!}$$

3. **Growth of $g(n)$**:
The common denominator can be computed by considering the least common multiple, which is approximately $(n!)^n$ particularly for large $n$. Thus, $g(n)$ can grow substantially, approximated using factorial growth:

$$n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n.$$

Hence,

$$(n!)^n \approx \left(\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\right)^n.$$

4. **Comparison with $n^{0.999n}$**:
We consider $n^{0.999n}$ and $g(n) \approx (n!)^n$. The factorial grows faster than the polynomial:

$$g(n) \approx (n!)^n \gg n^{0.999n}.$$

5. **Existence of $n$ with $g(n) > n^{0.999n}$**:
As factorial growth is much more rapid than the polynomial given, there exists an $n$ such that

$$g(n) = \Theta((n!)^n) > n^{0.999n}.$$

Thus, there indeed exists such an $n$.

The conclusion is that there exists positive integers $n$ for which $g(n) > n^{0.999n}$, thus:

$$\boxed{\text{yes}}$$