jmc

Suppose $a,$ $b,$ $c$ are three consecutive terms of the sequence. Then $b=ac-1,$ so $$c=\frac{b+1}{a}.$$Let $a_n$ denote the $n$th term. Then $a_1=x,$ $a_2=2000,$ and $$\begin{array}{rl}{a}_3& =\frac{a_2+1}{a_1}=\frac{2001}{x},\\ a_4& =\frac{a_3+1}{a_2}=\frac{2001/x+1}{2000}=\frac{x+2001}{2000x},\\ a_5& =\frac{(x+2001)/(2000x)+1}{2001/x}=\frac{x+1}{2000},\\ a_6& =\frac{(x+1)/2000+1}{(x+2001)/(2000x)}=x,\\ a_7& =\frac{x+1}{(x+1)/2000}=2000.\end{array}$$Since $a_6=a_1$ and $a_7=a_2,$ and each term depends only on the previous two terms, the sequence becomes periodic from this point, with period 5. Therefore, the first five terms represent all possible values.

We can have $a_1=x=2001.$

We can have $a_3=\frac{2001}{x}=2001,$ which leads to $$x=1.$$We can have $a_4=\frac{x+2001}{2000x}=2001,$ which leads to $$x=\frac{2001}{4001999}.$$We can have $a_5=\frac{x+1}{2000}=2001,$ which leads to $$x=4001999.$$Thus, there are $\boxed{4}$ different possible values of $x.$