14. (15 points) Let $n$ be a positive integer. If from any $n$ non-negative integers, one can always find four distinct numbers $a, b, c, d$ such that $a+b-c-d$ is divisible by 20, what is the minimum value of $n$?
【Analysis】First, explain that any 8 non-negative integers cannot satisfy the condition, because taking any 9 non-negative integers, and choosing any 7 from them, their pairwise sums result in 21 sums, and the remainders of these 21 sums when divided by 20 have 21 values. Since the remainders can have at most 20 different values, one of the following two scenarios must occur: (1) There are 4 different numbers $a, b, c, d$ such that $a+b$ and $c+d$ have the same remainder when divided by 20, in which case the four numbers satisfy the condition. (2) There are 3 different numbers $a, c, x$ such that $a+x$ and $b+x$ have the same remainder when divided by 20, then $(a+x)-(c+x)=a-c$ is a multiple of 20, and this can be repeated, thus solving the problem.

【Solution】Solution: There exist 8 numbers: $0,1,2,4,7,12,20,40$ such that no four of them can satisfy the condition, so the minimum value of $n$ is at least 9.

Because taking any 9 non-negative integers, and choosing any 7 from them, their pairwise sums result in 21 sums, and the remainders of these 21 sums when divided by 20 have 21 values. Since the remainders can have at most 20 different values, one of the following two scenarios must occur:
(1) There are 4 different numbers $a, b, c, d$ such that $a+b$ and $c+d$ have the same remainder when divided by 20, in which case the four numbers satisfy the condition.
(2) There are 3 different numbers $a, c, x$ such that $a+x$ and $b+x$ have the same remainder when divided by 20, then $(a+x)-(c+x)=a-c$ is a multiple of 20. Taking $a, c$ out, and from the remaining 7 numbers, similarly, we get: either four different numbers satisfy the condition, or two numbers $b, d$ such that $b-d$ is a multiple of 20. In this way, there are always $a, b, c, d$ such that $a+b-c-d$ is divisible by 20.
In summary, the minimum value of $n$ is 9.