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, $S=-2000$.
Best Answer
Hint :
$$ \lfloor a \rfloor + \lfloor a + \frac{1}{2} \rfloor = \lfloor 2a \rfloor $$
This follows from Hermite's identity.