Solution. Let $S_{n}$ be the set of all numbers $\left|a_{i}-a_{j}\right|, 1 \leq i<j \leq n-1, n \geq 3$. Then $a_{n}$ is the smallest number in the set $S_{n}$. Since $S_{n-1} \subset S_{n}$, then $a_{n-1} \geq a_{n}$, i.e., starting from $a_{3}$, the sequence is non-increasing. Therefore, $a_{9} \geq a_{10}=1$. Further, from the definition of the sequence, it follows that $a_{10} \leq\left|a_{8}-a_{9}\right|=a_{8}-a_{9} \Rightarrow a_{8} \geq a_{9}+a_{10} \geq 2$. Similarly, $a_{7} \geq a_{8}+a_{9} \geq 2+1=3$, $\quad a_{6} \geq a_{7}+a_{8} \geq 3+2=5, \ldots, a_{3} \geq a_{4}+a_{5} \geq 8+13=21$. This value is achieved if we take $a_{1}=55, a_{2}=34$.

Criteria.

(-.) It is strictly proven that $a_{n} \geq a_{n+1}$ for $n \geq 3$.

$(-/+)$ Answer + incomplete example: it is stated that $\min \left(a_{3}\right)=21$, and for this case, all terms of the sequence except $a_{1}, a_{2}$ are written out,

$(+/ 2)$ answer + example (including $a_{1}, a_{2}$), justification of minimality is absent or contains gross errors (for example, it is stated that in any such sequence, each number starting from the 3rd is equal to the difference of the two preceding ones,

or

it is proven that $a_{3} \geq 21$, but without proof, it is used that $a_{n} \geq a_{n+1}$, an example of a sequence for $a_{3}=21$ is absent, or

answer + incomplete example (without $\left.a_{1}, a_{2}\right)+$ significant progress in justifying minimality.

(+/-) It is strictly and without gaps proven that $a_{3} \geq 21$, an example is absent, or

answer + example (including $a_{1}, a_{2}$) + justification of minimality with minor gaps.

(+.) Logically correct and complete solution with no significant gaps except the following

- incomplete example (not explicitly written $a_{1}, a_{2}$),
- the fact that $a_{n} \geq a_{n+1}$ for $n \geq 3$ is used without proof,
- an arithmetic error is made in calculating $a_{3}$.

(+) It is strictly and without gaps proven that $a_{3} \geq 21$, and $a_{1}, a_{2}$ are explicitly written out, for which $a_{3}=21$.
9.6. The sequence $a_{n}$ is constructed as follows: $a_{1}, a_{2}$ are arbitrary real numbers, and for $n \geq 3$, the number $a_{n}$ is the smallest of the numbers $\left|a_{i}-a_{j}\right|, 1 \leq i<j \leq n-1$. For example, if $a_{1}=6, a_{2}=\frac{19}{2}$, then we get the sequence $6, \frac{19}{2}, \frac{7}{2}, \frac{5}{2}, 1,1,0,0,0 \ldots$ For some choice of $a_{1}$ and $a_{2}$, the sequence resulted in $a_{10}=1$. Find the smallest possible value of $a_{3}$ in such a sequence.