Let the nine integers be $x_{1,\ldots,9}$. There are $^9C_2=36$ pairs $\{i,j\}$ so by the pigeonhole principle there's a $t$ such that two pairs index equal sums: $x_a+x_b=x_i+x_j=t$. If $(a,b,i,j)$ are distinct, we're done. Otherwise, WLOG $(a,b,j)$ are distinct and $i=a$, thus $x_b=x_j$. Set $\{b,j\}$ aside and look now at the $^7C_2=21$ pairs remaining. Pigeonholing again, we can find $(a^\prime,b^\prime,i^\prime,j^\prime)$ similar to last time, either all distinct (in which case, we're done), or s.t. wlog $x_{b^\prime}=x_{j^\prime}$. But in that final case, we have $x_b+x_{b^\prime}=x_j+x_{j^\prime}$ $\quad\blacksquare$

jasperE3 wrote: Given nine positive integers, is it always possible to choose four different numbers $a,b,c,d$ such that $a+b$ and $c+d$ are congruent modulo $20$? Let $R=\{r_1, r_2, r_3,... r_9\}$ be collection of $9$ positive integers such that $r_1\leq r_3\leq...\leq r_9$, $0\leq r_i\leq 19$ for any $1\leq i\leq 9$ and we can't choose any $4$ numbers from $R$ such that sum of any pairs out of those $4$ is congruent same $\mod 20$ Claim 1-: If such $R$ exist then we can't have any $r_x,r_y\in R$ such that $x\neq y$ and both $r_x,r_y$ is repeated more then $1$ times. And also for any $1\leq k\leq 9$, $r_k$ is not repeated more then $3$ times. Prove-: Observe If such $r_x,r_y$ numbers exist then we will have $r_x+r_y\equiv r_x+r_y\mod 20$. Which is not possible. And if such $r_k$ exist in collection $R$ which is repeated more then $3$ times then also we can always choose $4$ $r_k$ from $R$ which is congruent to same $\mod 20$ when added in pairs. Which ka not possible Claim 2-: if Such $R$ exist and if $X(r_x, r_y)=r_x-r_y-1$ for $r_x>r_y$ Then we can't have $X(r_a, r_b)=X(r_c, r_d)$ for any $r_a, r_b, r_c, r_d\in R$ Prove-: Straight forward by definition of $R$ Claim 3-: There doesn't exist Any such $R$ Prove-: FTSOC assume such $R$ exist so it must follow Claim $1,2$ but Subcalim -: Following Claim $1,2$ we can only get atmost $6$ different $r_i$ from complete residue system $\mod 20$ even if we start from $r_1=0,r_2=1,r_3=2$ to get maximum possible number of $r_i's$. Prove-: Start from $r_1=0, r_2=1,r_3=2$ Then by claim $2\implies r_4=4$ Again Claim $2\implies r_5=7\implies$ Again By Claim $2\implies r_6=12$ Again Claim $2 \implies r_7=19$ but then $19+7\equiv 2+4\mod 20$ which is contradiction. Hence we can get atmost $6$ $r_i's$ Now by Claim $1$ We can get that $6$ different values of $r_i$ is Not sufficient. To prove the existence of $R$ Hence our assumption was wrong and Our claim $3$ is proved and we are done $\blacksquare$

do they really expect 3rd graders to do this?

Again, 3rd grade means the third grade of high school, i.e. 11th grade in most other systems.