Ответ. $44$ moves.

Let us prove that there must have been at least $20$ unpleasant moves (and thus the number of pleasant moves cannot exceed $44$). Let's place numbers in the cells as shown in рис. 3; cells with the same numbers are equidistant from the center, and cells with smaller numbers are closer to the center than those with larger numbers.

Рис. 3

Рис. 4

Every move from a cell with number $1$ does not decrease the distance to the center and is therefore unpleasant — there are $4$ such moves. A move from a cell with number $2$ can be pleasant only if it goes to a cell with number $1$. But there are eight cells with number $2$ and only four with number $1$, so at least four moves from cells with number $2$ will be unpleasant.

Now consider moves leading to the $32$ cells with numbers not less than $6$. Note that these moves cannot originate from cells with numbers $1$ or $2$, meaning they weren't accounted for in the previous reasoning. Such a move can only be pleasant if it comes from a cell with a number not less than $7$; however, there are only $20$ such cells. Therefore, among these moves, there are at least $32 - 20 = 12$ unpleasant ones, bringing the total number of unpleasant moves to no fewer than $4 + 4 + 12 = 20$.

An example of a traversal with $44$ pleasant moves is shown in рис. 4.