Let $f^r$ denote $\underbrace{f\circ f\circ f\circ \cdots \circ f}_r$. Since $f(n)\equiv 2n\pmod{1995}$, $f^r(n)\equiv 2^rn\pmod{1995}$. Then if $f^r(n)$ is divisible by 1995, so is $2^rn$. But 1995 and 2 are relatively prime, so 1995 must divide $n$.

If $n$ is a multiple of 1995, then so is $f^r(n)$ for any $r$. Also, $\frac{a}{1000}-1< \left\lfloor \frac{a}{1000} \right\rfloor \leq \frac{a}{1000}$, so $\frac{a}{200}\leq f(a)< \frac{a}{200}+1995$ for any positive integer $a$. Suppose there exists a positive integer $n$ which is a multiple of 1995 but $f^r(n)\neq 1995$ for any $n$. Let $m$ be the least positive value of $f^r(n)$. Then $m\geq 1995\cdot 2$ since $m$ is a multiple of 1995. But the inequality gives \[ 0<\frac{m}{200}\leq f(m)<\frac{m}{200}+1995\leq \frac{101m}{200}<m \] so $0<f(m)<m$ which is a contradiction. Therefore the sequence must eventually reach 1995. When $n=1997\cdot 500$, $\frac 52 \cdot 1995\leq f(n)<\frac 72 \cdot 1995$ so $f(n)$ must equal $3\cdot 1995$. Since $\frac{3\cdot 1995}{200}$ is clearly less than 1995, $f^2(n)=1995$, so $r=2$ in this case.