9.4. Vikentiy has two jars, a red one and a blue one, and a pile of 20 pebbles. Initially, both jars are empty. A move in Vikentiy's game consists of transferring a pebble from the pile to one of the jars or returning a pebble from one of the jars to the pile. The number of pebbles in the jars determines the game position. After each move, the number of pebbles in the red jar is always not less than the number of pebbles in the blue jar; and during the game, no position can be repeated. What is the maximum number of moves Vikentiy can make?
Answer: 110.

Solution: The position in Vikenty's game is uniquely defined by a pair of non-negative integers $(x, y)$, and $x+y \leq 20$, where $0 \leq y \leq x \leq 20$ are the numbers of pebbles in the blue and red jars, respectively. In total, there are $21+19+17 \ldots+1=121$ positions in Vikenty's game. We will call a position even if the sum of the numbers of pebbles in the jars is even, and odd otherwise. In total, there are $11+10+\ldots+1=66$ even positions and $121-66=55$ odd positions in the game. Each move in the game changes the parity of the position, so the game cannot contain more than 55 even and 56 odd positions (taking into account that it starts with an even position $(0,0)$), a total of 111 positions, and consist of no more than $111-1=110$ moves.

We will provide an example of a game where Vikenty can make 110 moves. In this case, he makes the only possible first move from $(0,0)$ to $(1,0)$, then he sequentially passes through all positions $(x, y)$ with odd $x=1,3,5, . ., 9$ from $(x, 0)$ to $(x, x)$, after which he moves to $(x+1, x)$, and all positions with even $x=2,4, . ., 10$ from $(x, x-1)$ to $(x, 0)$, after which he moves to $(x+1,0)$. After all of the above, he ends up in the position $(11,0)$, after which his strategy changes slightly. He sequentially passes through all positions $(x, y)$ with odd $x=11,13,15, .17$ from $(x, 0)$ to $(x, 19-x)$, after which he moves to $(x+1,19-x)$, and all positions with even $x=12,14, . ., 18$ from $(x, 20-x)$ to $(x, 0)$, after which he moves to $(x+1,0)$. In the end, he ends up in the position $(19,0)$ and makes the last move to $(20,0)$. The unvisited positions are $(2,2),(4,4), \ldots,(10,10),(11,9),(13,7), \ldots,(19,1)$ - exactly 10 of them, that is, exactly $121-10=111$ positions were visited. Exactly 110 moves were made.

Grading criteria. ($\cdot$) Proved that the number of moves is no more than 110: 3 points. ($\cdot$) Construction of an example with 110 moves with precise justification: 4 points. ($\cdot$) Lack of precise justification of the example: minus 2 points.