14. 3 piles of peaches have the numbers $93, 70, 63$, respectively. A monkey moves peaches between the 3 piles. It is known that the monkey can move a maximum of 5 peaches each time, and will eat 1 peach during the move from one pile to another. When the number of peaches in the 3 piles is equal, the monkey has eaten at least $\qquad$ peaches.
【Analysis】First, calculate the total number of peaches as $93+70+63=226$. Since they are divided into 3 piles with equal numbers, divide 226 by 3 and make the corresponding calculations based on the result.

【Solution】Solution: $93+70+63=226$,
$$
226 \div 3=75 \cdots 1 \text {, }
$$

Therefore, the number of peaches eaten by the monkey is $1, 4, 7$, etc.
If 1 peach is eaten, the monkey has moved 1 time, but moving 1 time cannot achieve equal numbers of peaches in the 3 piles. If 4 peaches are eaten, the monkey has moved 4 times, then the remaining number of peaches is $226-4=222$,

The number in each pile is $222 \div 3=74$,
93-74=19, so peaches can be moved from the first pile to the second and third piles. Move 1 time to the second pile, 5 peaches are moved and 1 is eaten, leaving 4, at this point the second pile has 74 peaches,

Then move 3 times to the third pile, each time moving 5, 5, and 4 peaches, leaving 4, 4, and 3 peaches respectively, at this point the first, second, and third piles all have 74 peaches, satisfying the problem's requirements.

Therefore, the answer is: 4.