Split two peas in a pod

We claim that $N=379$ . Note that for $N=378$ , then $S_{378} = \left\{1, 2, \cdots , 378\right\}$ . Setting $B=\left\{19, 20, \cdots , 360\right\}$ and the rest in $A$ , we can see that if $x_1, x_2, \cdots, x_{19}$ are all in $A$ , then either $19 \leq x_1+x_2+\cdots+x_{19} \leq 342$ or $x_1+x_2+\cdots+x_{19} \geq 379$ . Hence there are no $x_1, x_2, \cdots, x_{20}$ in $A$ satisfying $x_1+x_2+\cdots+x_{19}=x_{20}$ .

Similarly, if $x_1, x_2, \cdots, x_{19}$ are all in $B$ , then $x_1+x_2+\cdots+x_{19} \geq 361$ . Hence there are no $x_1, x_2, \cdots, x_{20}$ in $B$ satisfying $x_1+x_2+\cdots+x_{19}=x_{20}$ .

Thus, $S_{378}$ can be split into two sets $A$ and $B$ without the condition being fulfilled. It is easy to see that for all $N\leq 378$ , $S_N$ can also be split into two sets $A$ and $B$ without the condition being fulfilled. Hence $N>378$ .

For $N=379$ , suppose we can split $S_{379}$ into two sets $A$ and $B$ such that there doesn't exist a set from which we can choose 20 elements $x_1, x_2, \cdots, x_{20}$ satisfying the given equation.

WLOG assume $1 \in A$ . It follows that $19 \in B$ . Hence $361 \in A$ . If $20 \in A$ then selecting $x_1=1$ , $x_2=x_3=\cdots=x_{19}=20$ , and $x_{20}=361$ yields a contradiction, since $x_1,x_2,\cdots,x_{20}$ are all in $A$ and satisfies the equation. Hence $20 \in B$ .

If $379\in A$ , then selecting $x_1=361$ , $x_2=x_3=\cdots=x_{19}=1$ , and $x_{20}=379$ yields a contradiction, since $x_1,x_2,\cdots,x_{20}$ are all in $A$ and satisfies the equation. Thus $379\in B$ . But then selecting $x_1=19$ , $x_2=x_3=\cdots=x_{19}=20$ , and $x_{20}=379$ yields a contradiction, since $x_1,x_2,\cdots,x_{20}$ are all in $B$ and satisfies the equation. Therefore for $N=379$ , no matter how we split $S_{379}$ into two sets $A$ and $B$ , the condition is always fulfilled.

Therefore, $N=379$ is the minimum value for $N$ for which the condition will always be fulfilled.

Suppose $N=19^2+18=379$ . WLOG $1\in A$ . If $x_1,...,x_{20}$ don't exist, then $19\in B$ , so $19^2\in A$ , and $19^2+18\in B$ . However, 20 could not be in either set, because $19^2 = 1+18(20)$ and $19^2+18 = 19+18(20)$ .

On the other hand, if $N=19^2+17=378$ then no $x_1,...,x_{20}$ exist for $A=\{1, ..., 18, 19^2, ... , 19^2+17\}$ and $B=\{19, ... , 19^2-1\}$ , which shows that $379$ in minimal.

We claim that $N = 379$ is the answer.

1. Suppose $N = 378$ satisfies the condition, then let $A =$ { $19, 20, 21, ..., 360$ } and $B =$ { $1, 2, ..., 18, 361, 362, ..., 378$ }. It is clear that we can't find the required $20$ numbers in both sets.

2. On the other hand, $N = 379$ satisfies the condition because, if not, let $1 \in B$ . Then $19 \in A$ . (because $19 = 19 \times 1$ .) Then $361 \in B$ . ( $19 + 19 + ... + 19$ ( $19$ times) $= 361$ , so $19$ and $361$ can't be in the same set.). For similar reasons, $361 + 18(1) = 379 \in A$ . Consider $20$ . If $20 \in A$ , then $18(20) + 19 = 379$ , a contradiction to no such numbers exist. Else, if $20 \in B$ , then $18(20) + 1 = 361$ , also a contradiction. Thus, we conclude that $N = 379$ is the minimum possible value that satisfies the requirements.

Suppose that $N \ge 379$ and that there is no way of splitting $S_N$ as desired. For definiteness, suppose that $1 \in A$ . Then $A$ cannot contain $19 = 19\times1$ , and hence $19 \in B$ . Then $B$ cannot contain $361=19\times19$ , and hence $361 \in A$ . Then $A$ cannot contain $379 = 361 + 18\times1$ , and hence $379 \in B$ . Since $379 = 18\times20 + 19$ , we deduce that $B$ cannot contain $20$ , and hence $20 \in A$ . But this is impossible, since $1,20,361 \in A$ and $361 = 18\times20+1$ .

Thus we deduce that $S_N$ can be split as required for any $N \ge 379$ .

If $N \le 378$ , then the decomposition $\begin{array}{rcl}A & = & \{1,2,\ldots,18,361,362,\ldots,378\}\cap S_N \\ B & = & \{19,20,\ldots, 360\}\cap S_N \end{array}$ shows that it is not always possible to split $S_N$ as desired in this case. Thus the smallest possible value of $N$ that we want is $379$ .