11. If from any $\mathrm{n}$ numbers, one can always select 4 numbers such that their sum is a multiple of 4, then what is the minimum value of $n$?
【Answer】7
【Explanation】Pigeonhole principle, residue classes.
Natural numbers divided by 4 have 4 residue classes, with remainders of $0,1,2,3 \ldots$ First, if any class has 4 or more numbers, their sum will be a multiple of 4, so each class can have at most 3 numbers; and $1+3=2+2$, so if there are 2 numbers with a remainder of 1, or more than 2 numbers with a remainder of 2, a multiple of 4 can certainly be formed. To avoid a multiple of 4, these two classes together can have at most 3 numbers,
(1) Remainders of $1,1,3$ or $3,3,1$; since 1 and 3 can form 4, the number of remainders of 0 can be at most 1, otherwise $0+0+1+3=4$, which is a multiple of 4. Similarly, the number of remainders of 2 can also be at most 1, in this case, the maximum number of numbers that are not multiples of 4 is $3+1+1=5$ (numbers)
(2) 3 remainders of 1, or 3 remainders of 3, can have 3 remainders of 0, but no remainders of 2, so in this case, the maximum number of numbers that are not multiples of 4 is:3+3=6 (numbers), by the worst-case principle, to ensure a multiple of 4, the minimum number is 3+3+1=7 (numbers)