Find the function!

For all $y$ in the range of $f,$ we have $yf(y) = 1,$ so $f(y) = 1/y.$ Since $999$ is in the range of $f,$ we have $f(999) = 1/999,$ so $1/999$ is in the range of $f.$ Then the range of $f$ contains $\left[\frac1{999},999\right]$ by the Intermediate Value Theorem , so for all $y$ in $\left[\frac1{999},999\right],$ $f(y) = 1/y.$ Hence statements 1,2,4 are all true.

On the other hand, let $y=g(x)$ be the equation of the line through the points $(999,1/999)$ and $(1000,999),$ and consider the function $f(x) = \begin{cases} 999 & \text{if } x \le 1/999 \\ 1/x & \text{if } \frac1{999} \le x \le 999 \\ g(x) & \text{if } 999 \le x \le 1000 \\ 999 & \text{if } x \ge 1000 \end{cases}$ Then the range of $f$ equals $\left[\frac1{999},999\right],$ and $f$ is continuous, so it satisfies the conditions of the problem. This shows that statements 3,5,6,7 are not necessarily true. Hence the answer is $\fbox{8}.$