【Answer】 667

【Solution】Through trial and error, we find that only $a_{3}, a_{6}, a_{9}, \ldots \ldots, a_{999}$ are not palindromic strings, while all others are. Therefore, we can directly obtain the answer: there are only 333 non-palindromic strings, and the remaining $1000-333=667$ are palindromic strings.

Next, we will provide a rigorous proof (this part can be ignored if there is no time during the exam):
Assume $a_{n}=P, a_{n+1}=Q$, then $a_{n+2}=Q \bar{P}, a_{n+3}=Q \bar{P} \bar{Q}$. Since $Q$ and $\bar{Q}$ on both sides of $\bar{P}$ can ensure the palindromic property, whether $a_{n+3}$ is a palindromic string depends on the situation of $\bar{P}$. If $a_{n}=P$ is a palindromic string, then $a_{n+3}=Q \bar{P} \bar{Q}$ is also a palindromic string; if $a_{n}=P$ is not a palindromic string, then $a_{n+3}=Q \bar{P} \bar{Q}$ is also not a palindromic string. Considering that $a_{1}$ and $a_{2}$ are both palindromic strings, $a_{4}, a_{7}, \cdots, a_{1000}$ and $a_{5}, a_{8}, \cdots, a_{998}$ are all palindromic strings. Since $a_{3}=B A$ is not a palindromic string, $a_{6}, a_{9}, \cdots, a_{999}$ are not palindromic strings. Thus, we have proven the initial guess.
4. In a sequence of English letter strings, the first letter string $a_{1}=A$, the second letter string $a_{2}=B$, and each subsequent letter string $a_{n}(n \geq 3)$ is formed by appending the reverse of $a_{n-2}$ to $a_{n-1}$. For example, $a_{3}=a_{2} \bar{a}_{1}=B A$ (we use $\overline{a_{i}}$ to denote the reverse of $a_{i}$, which is reading the letter string from right to left, such as $\overline{A B B}=B B A$, $\overline{A A B A}=A B A A$), $a_{4}=a_{3} \overline{a_{2}}=B A B$, $a_{5}=a_{4} \overline{a_{3}}=B A B A B$, $a_{6}=a_{5} \overline{a_{4}}=B A B A B B A B$. Therefore, among the first 1000 letter strings in this sequence, there are $\qquad$ palindromic letter strings (a palindromic letter string is one that reads the same from left to right as from right to left, such as $A B A$, $A A B A A$)