Four, (This question is worth 30 points) Let the function $f(x)=\ln x+a\left(\frac{1}{x}-1\right), a \in \mathbb{R}$, and the minimum value of $f(x)$ is 0,
(1) Find the value of $a$; (2) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=f\left(a_{n}\right)+2\left(n \in \mathbb{N}_{+}\right)$, let $S_{n}=\left[a_{1}\right]+\left[a_{2}\right]+\left[a_{3}\right]+\cdots+\left[a_{n}\right]$, where $[m]$ denotes the greatest integer not exceeding $m$. Find $S_{n}$.
Solution: (1) $f^{\prime}(x)=\frac{1}{x}-\frac{a}{x^{2}}=\frac{x-a}{x^{2}}, x>0$.
When $a \leqslant 0$, $f^{\prime}(x)>0$, then $f(x)$ is monotonically increasing on $(0,+\infty)$, with no minimum value, which does not meet the problem's requirement.
When $a>0$, for $0<a$, then $f^{\prime}(x)<0$; for $x>a$, then $f^{\prime}(x)>0$. Therefore, the function $f(x)$ is monotonically decreasing on $(0, a)$ and monotonically increasing on $(a,+\infty)$.
Thus, $f(x)_{\text {min }}=f(a)=\ln a-a+1$.
Let $g(a)=\ln a-a+1(a>0)$. Then $g^{\prime}(a)=\frac{1}{a}-1=\frac{1-a}{a}$.
If $0<a<1$, then $g^{\prime}(a)>0$; if $a>1$, then $g^{\prime}(a)<0$. Therefore, $g(a)$ is monotonically increasing on $(0,1)$ and monotonically decreasing on $(1,+\infty)$.
Thus, $g(a)_{\text{max}}=g(1)=0$, and $g(a)<0$ for $a \neq 1$.
Since $g(2)=\ln 2-1<0$, $g(3)=\ln 3-2<0$, and $g(2)<g(3)$, we have $2<h(a_4)<3$, i.e., $2<a_{i+1}<3$.
Thus, when $n=k+1$, the conclusion also holds.
From (I) and (II), for all integers $n \geqslant 3$, we have $2<a_{n}<3$.
Therefore, $\left[a_{1}\right]=1,\left[a_{n}\right]=2(n \geqslant 2)$.
Hence, $S_{n}=\left[a_{1}\right]+\left[a_{2}\right]+\left[a_{3}\right]+\cdots+\left[a_{n}\right]=1+2(n-1)=2 n-1$. 30 points