The expression $f(x):=\left(1+\frac{1}{x}\right)^{x+1}$ may not be a real number for $x\in[-1,0]$. However, for $x\in (-\infty,-1)$, $f(x)$ increases to $\text{e}$ as $x$ decreases to $-\infty$. Hence, for $x<-1$ and $t\in\mathbb{R}$, $f(x)=t$ has at most one solution, and there exists exactly one solution if and only if $0<t<\text{e}$. From user90369's comment, we see that $x=-2000$ is the unique solution to $f(x)=\left(1+\frac{1}{1999}\right)^{1999}$ with the extra condition that $x<-1$.
Next, for $x>0$, $f(x)$ decreases to $\text{e}$ as $x$ increases to $+\infty$. Hence, for $x>0$ and $t\in\mathbb{R}$, $f(x)=t$ has at most one solution, and there exists exactly one solution if and only if $t>\text{e}$. This proves that $f(x)=\left(1+\frac{1}{1999}\right)^{1999}$ has no solution with $x>0$.
Now, let $x\in(-1,0)$. Only for $x=-\frac{p}{q}$ where $p,q\in\mathbb{Z}_{>0}$, $\gcd(p,q)=1$, and $q$ is odd, we have that $f(x)$ is real. For such $x=\frac{p}{q}$, $f(x)>0$ if and only if $p$ is odd. That is, if $f(x)=\left(1+\frac{1}{1999}\right)^{1999}$, then $x=-\frac{2m-1}{2n-1}$ for some integers $m,n>0$ (with $m<n$ and $\gcd(2m-1,2n-1)=1$). However, we obtain
$$f\left(-\frac{2m-1}{2n-1}\right)=\left(\frac{2(n-m)}{2m-1}\right)^{ \frac{2(n-m)}{2n-1}}\,.$$
Note that this number is a rational number if both $2(n-m)$ and $2m-1$ are perfect $(2n-1)$-st powers of integers. Since $2(n-m)=b^{2n-1}$ for some integer $b>0$ and $2(n-m)>1$, we get $b\geq 2$, whence
$$2n-1>2(n-m)=b^{2n-1}\geq 2^{2n-1}>2n-1\,,$$ which is absurd. That is, for $x\in(-1,0)$ such that $f(x)$ is a positive real number, $f(x)$ is an irrational number. Hence, $f(x)=\left(1+\frac{1}{1999}\right)^{1999}$ has no solution with $x\in[-1,0]$ either.
In conclusion, the only real solution to $f(x)=\left(1+\frac{1}{1999}\right)^{1999}$ is $x=-2000$. That is, $S=-2000$.
Let's start with a simple proof showing that when including negative numbers, it's possible for $\bigg[\frac{[x]}{x}\bigg]$ to be any natural number larger than or equal to 1.
Let $x=-\frac{1}{a}$ and $a>1$, then $-1<x<0$. Since $x$ is always between $-1$ and $0$ (in this specific example defining $x$ by $a$), then the greatest integer less than $x$ is always $-1$:
$$\implies[x]=-1$$
So forth, plugging this into the formula
$$\frac{[x]}{x}=\frac{-1}{-\frac{1}{a}}=a$$
Thus, $\forall a>1$ there exists an $x\in(-1,0)$, such that $\frac{[x]}{x}=a$.
So we've shown that $\frac{[x]}{x}$ has range at least $\mathbb{R}_{>1}$, which implies $\bigg[\frac{[x]}{x}\bigg]$, has range of at least $\mathbb{N}_{\geq1}$.
But since $\frac{[x]}{x}\geq0$ $\forall x$ and as you yourself have shown that when $x>0$ then $\bigg[\frac{[x]}{x}\bigg]$ has range $\{0,1\}$ (E.g; $\bigg[\frac{[0.5]}{0.5}\bigg]=0$ and $\bigg[\frac{[1]}{1}\bigg]=1$), then the entire range is $\mathbb{N}_{\geq1}\cup\{0,1\}=\mathbb{N}_0$.
In summary, when $x<0$, we find that $\bigg[\frac{[x]}{x}\bigg]$ can be any natural number greater than or equal to $1$, and when $x>0$ we find that $\bigg[\frac{[x]}{x}\bigg]$ can (and must) be $0$ or $1$, and since $\bigg[\frac{[x]}{x}\bigg]$ as a whole is always greater than or equal to $0$, then the range is all natural numbers including zero.
A good question and a fun solution!
Best Answer
Here are some hints.
Note that as $x$ increases, the sum can get no larger - it is decreasing but not strictly so, because it is sometimes constant. Also with $x=1$ the sum is equal to $7$ so we must have $x \gt 1$ for a sum as low as $5$.
Now if $x\gt 1$ we have immediately that $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]\le 2+3=5$ so we must have equality with $\left[\frac{3}{x}\right]=2$ and $\left[\frac{4}{x}\right]=3$
You should be able to complete things from there.