The maximum is 985. To prove this, we start with two lemmas.

Lemma 2. Suppose we begin with the pair of integers $(1, 1)$, and each step we are allowed to replace one of the integers with the sum of both. Then after $k$ steps, the maximum possible sum of the two numbers is $F_{k+3}$, where $F_n$ is the $n$th Fibonacci number with $F_0 = 0$, $F_1 = 1$.

Proof. Let the numbers be $a, b$ with $a < b$. We show by induction that $a \le F_{k+1}$ and $b \le F_{k+2}$ after $k$ steps. The base case is given. Given it is true for $k$, suppose a pair we get after $k+1$ steps comes from the pair $a, b$. The new pair is either $a, a+b$ or $b, a+b$. By the induction hypothesis $a+b \le F_{k+1}+F_{k+2} = F_{k+3}$, and we already know that both $a$ and $b$ are at most $F_{k+2}$. This completes the induction and the proof. $\square$

Lemma 3. Let $\frac{a}{b}$ and $\frac{c}{d}$ be two rational numbers in lowest terms with $0 \le \frac{a}{b} < \frac{c}{d} \le 1$ which are adjacent terms of a Farey series; that is, there does not exist $\frac{p}{q}$ with $\frac{a}{b} < \frac{p}{q} < \frac{c}{d}$ and $q < \max(b, d)$. Then of all fractions $\frac{p}{q}$ with $\frac{a}{b} < \frac{p}{q} < \frac{c}{d}$, $\frac{a+c}{b+d}$ has the smallest denominator.

Proof. We use a lattice point interpretation. The condition of $\frac{a}{b}, \frac{c}{d}$ being adjacent in a Farey series means that the triangle with vertices $(0,0)$, $(a,b)$, $(c,d)$ has area $\frac{1}{2}$, which is a consequence of Pick's Theorem. Now notice that in the triangle $(0,0)$, $(na,nb)$, $(nc,nd)$, for any nonnegative integers $k, m$ with $k+m \le n$ we have $(ka+mc, kb+md)$ in or on the triangle. We also know the triangle's area is $\frac{n^2}{2}$. Combining these two observations with Pick's Theorem shows that the points we have described are in fact all of the points inside or on the triangle. Using this and taking $n$ arbitrarily high, we conclude that if $\frac{p}{q}$ is a fraction with $\frac{a}{b} < \frac{p}{q} < \frac{c}{d}$, then there exist positive integers $k, m$ with $p = ka + mc$ and $q = kb + md$. Note $\text{gcd}(p, q) = 1$ if $\text{gcd}(k, m) = 1$. The desired conclusion is immediate. $\square$

Now we prove our main result. Note that if $a_1, a_2, \dots, a_n$ is a regular sequence, then $a_1 + m, a_2 + 2m, \dots, a_n + nm$ is also regular by replacing the associated $x$ with $x+m$. Therefore we may assume $a_1 = 0$ and $0 \le x < 1$ in the regularity condition.

For each finite regular sequence, we can associate two real numbers $r, s$ to it so that the real numbers $x$ which satisfy the condition for being regular are those with $r \le x < s$. We claim $r, s$ are always rational numbers which are adjacent terms of a Farey series. We show this by induction on the number of terms. For a 1-term regular sequence with $a_1 = 0$, we have $r = 0$ and $s = 1$. Given that we have it is true for all $n$-term regular sequences, note that if there is no integer multiple of $\frac{1}{n+1}$ between $r$ and $s$, then $a_{n+1}$ is forced and $r, s$ remain the same as before. Otherwise, if $r < \frac{k}{n+1} < s$, then $a_{n+1}$ can be either $a_n$ or $a_n + 1$. If $a_{n+1} = a_n$, we can check the new $r, s$ will be $r, \frac{k}{n+1}$. If $a_{n+1} = a_n + 1$, it will be $\frac{k}{n+1}, s$. The inductive hypothesis tells us that in both cases we obtain adjacent terms of a Farey series.

Combining the above observations with lemma 3, we can analyze forced terms as follows: if at a given point we have $r = \frac{a}{b}$ and $s = \frac{c}{d}$, then the next time we will have an unforced term is at step $b+d$, when one of $r$ and $s$ will change to $\frac{a+c}{b+d}$ depending on the choice of the $b+d$th term. In particular, the denominators of $r, s$ start at $(1,1)$, and each time we have an unforced term, one denominator becomes the sum of the previous two. Since our sequence has 1000 terms, our goal is to have the sum of denominators greater than 1000 in as few unforced terms as possible, which would make the next unforced term come after the end of our sequence. By lemma 2 and the fact that $F_{16} = 987$, $F_{17} = 1597$, at least 14 unforced terms are required. Combining this with the fact that $a_1$, which we didn't count yet, is also unforced, we have at least 15 unforced terms, and thus at most 985 forced terms as claimed.

To construct a sequence with this many forced terms, we simply set
$$a_k = \lfloor k\varphi \rfloor \quad \text{for } 1 \le k \le 1000,$$
where $\varphi = \frac{\sqrt{5}-1}{2} \approx 0.618...$. This will make $r, s$ always take the form $\frac{F_k}{F_{k+1}}$ for some $k$ and obtain the equality case of lemma 2.