1, analysis: (1) First find out the position of the ship on the left bank, where odd-numbered ships cross the river (the ship is on the right bank), and where even-numbered ships cross the river (the ship is on the left bank), just use the rules found;

(2) Using the law discovered by (1), 20 10 times is even, and the problem can be solved directly by crossing the river several times.

3) The ship was originally on the left bank. As soon as it crossed the river, it reached the right bank, and as soon as it crossed the river, it returned to the left bank from the right bank, that is, it crossed the river twice. Therefore, when the ship leaves from the left bank and returns to the left bank many times, the number of crossing the river must be a multiple of 2, so it is even. In the same way, it is not difficult to draw that if the ship finally stops on the right bank, then the number of times it crosses the river must be odd.

Answer: solution: it is not difficult to find that if the ship is on the left bank at first, it will return to the left bank after several rivers; After crossing the river odd times, stop at the right bank. Now the boat crosses the river 20 10 times, even several times. Therefore, the ship should finally stop at the left bank.

Note: the key to solve this problem is the understanding of the number of crossing the river: one way, that is, crossing the river once from the left bank to the right bank (or from the right bank to the left bank); Cross the river twice round trip.

2, solution, from the meaning of the question,

n=n| 1| >|x 1|+|x2|+...|xn|= 19+|x 1+x2+...xn|≥ 19

So n≥20

So the minimum value of n is n=20.

x 1 = x2 =……= x 10 = 19/20

x 1 1 = x 12 =……= x20 =- 19/20

find