No more fault

Let there are $n,n\geq 4$ numbers $a_1,a_2,.....a_{n-1},a_n$ , all are distinct. Now if for four distinct numbers $a+c-b-d$ is divisible by 20 then ,

$|(a-b)(\text{mod} 20)|=|(c-d)(\text{mod} 20)|$ . Where the modulos vary from -9 to 10 over the integers. So, in sake of simplification we say that the order pair of two elements, let $(a_i,a_j)=|(a_i-a_j)(mod 20)|$ . And we shall work with this pairs from now on.

So, our goal is to prevent four numbers from satisfying the given condition ( $20|a-b+c-d)$ . So to do that we have put our maximum effort not to give any of those disjoint pairs ( $(a_p,a_q)=(a_k,a_l)$ , p,q,k,l all are distinct) to be equal, namely . Intersecting pairs may be equal because we are asked for four distinct numbers. So, we shall proceed like this.

Let, $0\leq (a_p,a_k)=r\leq 10$ . And let, all $r$ 's are used to prevent equality with the best effort. Without loss of generality we take $(a_1,a_2)=0$ . Now, if any of the values $(a_p,a_q)=0;p,q=3,4,....n-1,n$ , then we are failed (we will have for distinct numbers $a_1,a_2,a_k,a_l;k,l \in$ { $3,4,...n-1,n$ } such that $20|a_1-a_2+a_k-a_l$ .

So, to prevent them from being $0$ we should give them other $9$ values ( $1,2...10$ ). Now, there are $A=$ $\frac{(n-2)(n-3)}{2}$ numbers of pairs with no $a_1$ , $a_2$ .

Now, if we have two equal-valued disjoint pairs in $S=$ { $(a_p,a_q) ;p,q=3,4,....n-1,n$ } then we are failed. But if two intersecting $(a_p,a_q)=(a_p,a_k)$ pair with a common element $a_p$ are of equal values then $(a_p,a_q)~(a_p,a_k)=(a_q,a_k)=0$ or $2r$ ; as we had taken modulus ; $(a_q,a_k)$ surely may not be $0$ but $2r$ or $2r(mod 20)$ . Then we are failed again.

So, we have $\frac{(n-2)(n-3)}{2} > \bold {2*10-1=19} \Rightarrow \frac{(n-2)(n-3)}{2} \geq 20 \Rightarrow n-2 \geq 7 \Rightarrow n\geq 9$ . We can take two non-disjointpairs equal $\rightarrow$ at most "twice" because the equality of pairs does not imply the fail of our prevention as $(a_p,a_q)~(a_p,a_k)$ $\Rightarrow (a_q,a_k)=0$ or $2r$ . But an extra equality fails our prevention.

So, the least such $n$ is equal to $\bold 9$ .

If we have a set of numbers belonging to seven or more residue classes $\pmod{20}$ then, since $\binom{7}{2} = 21 > 20$ , we must be able to choose four distinct numbers $a,b,c,d$ , each with a different residue $\pmod{20}$ , such that $a + b \equiv c+d \pmod{20}$ .

If we have a set of numbers, and if four of them all belong to the same residue class $\pmod{20}$ , then these four numbers $a,b,c,d$ are distinct numbers such that $a+b\equiv c+d \pmod{20}$ .

If we have set of numbers, and if two pairs of numbers $a,c$ and $b,d$ exist such that $a$ and $c$ belong to the same residue class $\pmod{20}$ , and such that $b$ and $d$ belong to the same residue class $\pmod{20}$ , then $a+b \equiv c+d \pmod{20}$ .

Thus, if our set of numbers has any of

• 7 or more residue classes $\pmod{20}$ ,
• one residue class with four or more elements,
• two residue classes with two or more elements,

then we can find distinct integers $a,b,c,d$ in the set such that $a+b \equiv c+d \pmod{20}$ . If our set has $9$ or more elements, then at least one of the above options must happen. Since it is possible to have a set of $8$ elements comprising $6$ residue classes, with one class containing $3$ numbers, it may be possible to find a set with $8$ elements that does not have the pair/sum/divisibility property. Indeed, the numbers $0,1,2,4,7,12,20,40$ do the trick.

This makes the answer $\boxed{9}$ .