5. Find the smallest positive integer $n(n \geqslant 4)$, such that from any $n$ different integers, four different numbers $a$, $b$, $c$, $d$ can always be selected, satisfying $a+b-c-d$ is divisible by 20.
5. Since the remainder of any integer divided by 20 can only be $0,1,2, \cdots, 19$, for a set containing $k$ integers from $\{0,1,2, \cdots, 19 \mid$, there can be $C_{k}^{2}=\frac{1}{2} k(k-1)$ different integer pairs. If $-\frac{1}{2} k(k-1)>20$, i.e., $k \geqslant 7$, then there exist two different pairs $(a, b)$ and $(c, d)$ such that $a+b \equiv c+d(\bmod 20)$, i.e., $a+b-c-d$ is divisible by 20, and $a, b, c, d$ are all different (in fact, we already have $a \neq b, c \neq d$, if $a=c$, then $b=d(\bmod 20)$. Since $b, d \in\{0,1,3, \cdots, 19\}$, so $b=d$, which contradicts that $(a, b)$ and $(c, d)$ are different pairs).

Generally, consider a set of 9 integers. If at least 7 of these 9 integers have different remainders modulo 20, then by the previous analysis, there must exist 4 numbers $a, b, c, d$ such that $a+b-c-d$ is divisible by 20. If the 9 integers have at most 6 different remainders modulo 20, then there must exist 4 numbers $a, b, c, d$ such that $a \equiv b=c \equiv d(\bmod 20)$, or $a \equiv c, b \equiv d(\bmod 20)$ (otherwise, there would be at most $1+1+1+1+1+3=8$ numbers), thus $a+b-c-d$ is divisible by 20. Next, consider the set of 8 numbers $A=\{10,20, 40,1,2,4,7,12\}$. In fact, the remainders of these 8 numbers when divided by 20 are $0,0,0,1,2,4,7,12$, and these numbers have the property that each number is greater than the sum of the two smallest numbers, and the sum of any two numbers is less than 20. Let $a, b, c, d$ be 4 different numbers in $A$, and consider 20 and 40 as 0. Let $a$ be the largest of the 4 numbers, then $0<a-c-d<a+b-c-d \leqslant a+b<20$, so $a+b-c-d$ is not divisible by 20. Therefore, the smallest value of $n$ is 9.