346. A circle is divided into 6 equal sectors, in which the numbers $0,1,2,0,2,1$ (in that order) are placed. It is allowed to add the same number to two adjacent numbers in one move. Is it possible to achieve, in several such moves, that all 6 numbers in the sectors are equal?
$\triangle$ Let at some step the numbers in the sectors are in sequential order $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. We form the following sum:

$$
S=a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}
$$

After each move, this sum does not change, as each of the differences $a_{1}-a_{2}, a_{3}-a_{4}, a_{5}-a_{6}$ remains the same when both the minuend and subtrahend are increased by the same number; therefore, it is an invariant. However, in the initial position, $S=0-1+2-0+2-1=2$, and in the final position, when all six numbers are equal to the same number, $S=0$. Therefore, it is impossible to make all six numbers equal.

Answer: cannot.