jmc

From the given information, we can derive that $$\begin{array}{rl}f(x)& =f(2158-x)=f(3214-(2158-x))=f(1056+x)\\ & =f(2158-(1056+x))=f(1102-x)\\ & =f(1102-(1056+x))=f(46-x)\\ & =f(398-(46-x))=f(352+x).\end{array}$$It follows that $f(x)$ is periodic, whose period divides 352. This means that every value in the list $f(0),$ $f(1),$ $\dots ,$ $f(999)$ must appear among the values $$f(0),f(1),f(2),\dots ,f(351).$$The identity $f(x)=f(398-x)$ implies that every value in the list $f(200),$ $f(201),$ $\dots ,$ $f(351)$ must appear among the values $$f(0),f(1),\dots ,f(199),$$and the identity $f(x)=f(46-x)$ implies that every value in the list $f(0),$ $f(1),$ $\dots ,$ $f(22)$ must appear among the values $$f(23),f(24),\dots ,f(199).$$This implies that $f(23),$ $f(24),$ $\dots ,$ $f(199)$ capture all the possible values of $f(n),$ where $n$ is a positive integer.

Now, let $f(x)=\cos \left(\frac{360}{352}(x-23)\right),$ where the cosine is evaluated in terms of degrees. Then $$1=f(23)>f(24)>f(25)>\cdots >f(199)=-1,$$and we can verify that $f(x)=f(398-x),$ $f(x)=f(2158-x),$ and $f(x)=f(3214-x).$

Thus, the list $f(0),$ $f(1),$ $\dots ,$ $f(999)$ can have at most $199-23+1=\boxed{177}$ different values.