Dear @leonbloy, it's local Olympiad on informatics in IRAN in 2000.

@squirrel My guess is that the statement is "not [strictly] lower than [either of] its neighbors," i.e., its value is at least both of its neighbors. Although we should probably wait for OP...

Dear @squirrel, suppose in a third box we have number 13. in the forth box we have 4. in the second box we have 6. so we find third box as a maximum between two neighbors. so in 31 boxes that we know anything about the number except if we open the box, we want to find such a number with at most 11 opening.

@MounaMokhiab Are we assuming that the numbers in the boxes are $1,2,\ldots, 31$?

Dear @angryavian, it's very nice question. no the number is in arbitrary order, and just we know each number by opening the box.

@MounaMokhiab So, if the second, third, and fourth boxes each have $13$, it is ok to choose the third box as our desired box?

Dear @angryavian, no number is random.

okey ?

@MounaMokhiab My question was are the numbers all integers from 1 to 31, but in some random order?

yes, the number is random

so no box has the number 34, for example?

maybe has.

we have condition on number in each box

ok, so what if you open the second, third, and fourth boxes, and all of them have the number "17.4." are we done because the third box's number is not lower than the numbers of its neighbor boxes?

I say not lower. it means equal.

?

the box number maybe start at 1...31 maybe 300 to 9000 or anything.

I think we divide 31 box into 4 section.

1..9..16......31

Please answer my previous question about three consecutive boxes having the same number "17.4."

and open each box in n/4

so the 17.4 is answer. because 17.4 is not lower from two neighbor.

ok

what do u think?

I think we must partition the interval into [n/4]. so open the 1,9,16,23,31 box. keep the interval that has the maximum number. and half the interval and open the box in the half. keep the interval that has maximum number and repeat.