16. Let the function $f(x)=\ln x+a\left(\frac{1}{x}-1\right), a \in \mathbf{R}$, and the minimum value of $f(x)$ is 0.
(1) Find the value of $a$;
(2) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=f\left(a_{n}\right)+2\left(n \in \mathbf{N}_{+}\right)$, let $S_{n}=\left[a_{1}\right]+\left[a_{2}\right]+\cdots+\left[a_{n}\right]$, where $[m]$ denotes the greatest integer not exceeding the real number $m$, find $S_{n}$.
(1) $f^{\prime}(x)=\frac{1}{x}-\frac{a}{x^{2}}=\frac{x-a}{x^{2}}$, when $a>0$, $f(x)$ is monotonically decreasing on $(0, a)$ and monotonically increasing on $(a,+\infty)$, so $f(x)_{\min }=f(a)=\ln a+1-a=0 \Rightarrow a=1$.
(2) $f(x)=\ln x+\frac{1}{x}-1 \Rightarrow a_{n+1}=\ln a_{n}+\frac{1}{a_{n}}+1$,

Now we prove: $a_{n}<3$. When $n=2$, $a_{2}=2<3$, the conclusion holds;
Assume the conclusion holds for $n=k$, i.e., $a_{k}<3$, when $n=k+1$, $a_{k+1}=\ln a_{k}+\frac{1}{a_{k}}+1<\ln 3+\frac{1}{3}+1<3$, so the conclusion also holds for $n=k+1$. By the principle of mathematical induction, $a_{n}<3$ holds for all $n \in \mathbf{N}_{+}$.
Therefore, $S_{n}=1+\underbrace{2+2+\cdots+2}_{n-1 \uparrow}=2 n-1$.