Solution. Suppose that a suitable initial position and a sequence of 35 jumps exist, and number the squares of the chessboard according to the following scheme:

Moves of length one lead from an odd number to an even number and vice versa. Moves of length two always lead from an even number to another even number and from an odd number to another odd number. If the visited squares are denoted by $P_{1}, P_{2}, \ldots, P_{36}$, it follows from the above that among the four squares $P_{2}, P_{3}$, $P_{4}, P_{5}$, each of the numbers is represented exactly once (on $P_{2}$ and $P_{3}$ are different numbers with the same parity, and similarly on $P_{4}, P_{5}$ with the other parity). For the same reasons, each of the numbers is represented exactly once in the quadruples of squares $P_{4 k+2}, P_{4 k+3}, P_{4 k+4}, P_{4 k+5}$ for each $k \in\{0,1, \ldots, 7\}$. Among the numbers on the squares $P_{2}, P_{3}, \ldots, P_{33}$, each of the numbers is thus represented a total of 8 times.

The number 4 appears on the chessboard only 8 times, so none of the numbers on $P_{1}, P_{34}, P_{35}, P_{36}$ can be 4. The numbers on $P_{34}$ and $P_{35}$ have the same parity and are different (they are separated by a jump of length 2). Since 4 is not on either of them, they must both be odd. Then the number on square $P_{36}$ must be even, and the same even number must also appear on square $P_{1}$. Therefore, both must be the number 2.

The initial square must thus be one of those shaded on the left chessboard. However, this argument can also be repeated for the second numbering on the right, which is just a "rotation" of the first numbering. Since no square is shaded on both chessboards, we have reached a contradiction. Therefore, it is impossible to traverse the chessboard in the desired manner, regardless of the initial square chosen.