The answer is 985 . WLOG, by shifting $a_{1}=0$ (clearly $a_{1}$ isn't forced). Now, we construct regular sequences inductively using the following procedure. Start with the inequality
$$
\frac{0}{1} \leq x<\frac{1}{1} \text {. }
$$

Then for each $k=2,3, \ldots, 1000$ we perform the following procedure. If there is no fraction of the form $F=\frac{m}{k}$ in the interval $A \leq x<B$, then $a_{k}$ is forced, and the interval of possible $x$ values does not change. Otherwise, $a_{k}$ is not forced, and we pick a value of $a_{k}$ and update the interval accordingly.

The theory of Farey sequences tells us that when we have a stage $\frac{a}{b} \leq x<\frac{c}{d}$ then the next time we will find a fraction in that interval is exactly $\frac{a+c}{b+d}$ (at time $k=b+d$ ), and it will be the only such fraction.

So essentially, starting with $\frac{0}{1} \leq x<\frac{1}{1}$ we repeatedly replace one of the endpoints of the intervals with the mediant, until one of the denominators exceeds 1000; we are trying to minimize the number of non-forced terms, which is the number of denominators that appear in this process. It is not hard to see that this optimum occurs by always replacing the smaller of the denominators, so that the sequence is
$$
\begin{array}{l}
\frac{0}{1} \leq x<\frac{1}{1} \\
\frac{0}{1} \leq x<\frac{1}{2} \\
\frac{1}{3} \leq x<\frac{1}{2} \\
\frac{1}{3} \leq x<\frac{2}{5} \\
\frac{3}{8} \leq x<\frac{2}{5} \\
\frac{3}{8} \leq x<\frac{5}{13}
\end{array}
$$
and so on; we see that the non-forced terms in this optimal configuration are exactly the Fibonacci numbers. There are 15 Fibonacci numbers less than 1000, hence the answer $1000-15=985$
A finite sequence of integers $a_{1}, a_{2}, \ldots, a_{n}$ is called regular if there exists a real number $x$ satisfying
$$
\lfloor k x\rfloor=a_{k} \quad \text { for } 1 \leq k \leq n .
$$

Given a regular sequence $a_{1}, a_{2}, \ldots, a_{n}$, for $1 \leq k \leq n$ we say that the term $a_{k}$ is forced if the following condition is satisfied: the sequence
$$
a_{1}, a_{2}, \ldots, a_{k-1}, b
$$
is regular if and only if $b=a_{k}$.
Find the maximum possible number of forced terms in a regular sequence with 1000 terms.