Japan Mathematical Olympiad

$19$

If one writes the number in each square as shown in the figure below, the sum of the numbers written in the squares where the coins are placed cannot be increased by the operations. The number written in the square where the coin was initially placed is $64$ and there are three squares with $1$ written in them, eight squares with $2$ and six squares with $4$. Since the number written in each of the other squares is at least $8$, $64<1\cdot 3+2\cdot 8+4\cdot 6+8\cdot 3$ shows that the number of coins placed in the squares is always less than or equal to $3+8+6+3-1=19$.

On the other hand, one can place $19$ coins in the squares by performing the operations as follows. In the following operation $(x)$ on the square in $i$th row from the top and $j$th column from the left is denoted by "$x$ on $(i,j)$".

Do $d$ on $(1,4)$, $b$ on $(2,5)$, $a$ on $(5,2)$, $d$ on $(6,2)$, $d$ on $(4,3)$, $a$ on $(5,2)$, $d$ on $(5,4)$, $d$ on $(6,3)$, $d$ on $(6,5)$, $d$ on $(3,4)$, $d$ on $(4,3)$, $a$ on $(4,5)$, $d$ on $(5,5)$, $d$ on $(6,6)$, $d$ on $(5,4)$, $c$ on $(2,3)$, $a$ on $(5,6)$, $d$ on $(4,5)$, $d$ on $(3,4)$ in that order.

From the above, the answer is $19$.

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Alternative solution.

The proof that the number of coins placed in the squares is at most $19$ is the same as above. One can also place $19$ coins in the squares by performing the operations as follows.

Do $d$ on $(1,4)$, $d$ on $(2,3)$, $d$ on $(3,2)$, $d$ on $(4,3)$, $a$ on $(5,2)$, $d$ on $(6,2)$, $d$ on $(5,4)$, $d$ on $(6,3)$, $d$ on $(6,5)$, $d$ on $(3,4)$, $a$ on $(4,3)$, $d$ on $(5,3)$, $d$ on $(4,5)$, $d$ on $(5,4)$, $a$ on $(5,6)$, $d$ on $(6,6)$, $d$ on $(2,5)$, $d$ on $(3,6)$, $d$ on $(4,5)$, $a$ on $(5,6)$, $d$ on $(3,4)$, $b$ on $(4,3)$, $c$ on $(4,5)$ in that order.

From the above, the answer is $19$.