I need help with this competitive problem, the only hint I have in the book is that the problem can be solved with mathematical induction:
You have sums of natural numbers $x_1 + x_2 + \dots + x_k = y_1 + y_2 + \dots + y_l < k \cdot l$. Prove that you can "delete" some numbers (at least 2 but not all) so equality will stay. E.g. $1+1+2 = 1+1+2 < 9$ we can delete two 2s and we will have $1+1=1+1$.
I tried induction by $k$ and $l$ and also by $k\cdot l$ but didn't get anything.
First note that $k\ge2$ and $l\ge2$. Proceed by induction on $k+l$.
If $k+l=4$ then the sets are both $\{1,1\}$ or both $\{1,2\}$ and the result is obvious.
Without loss of generality we can suppose that $x_1 \le x_2 \le\dots \le x_k$ and $ y_1 \le y_2 \le \dots \le y_l $.If $x_k=y_l$ then we are finished. So we can suppose $x_k>y_l$. Deleting $y_l$ and replacing $x_k$ by $x_k-y_l$ keeps the two sums equal and (since we have removed the largest $y_i$) this sum is less than $k(l-1)$.
Then, by the inductive hypothesis two proper subsets of the $x_i$ and of the $y_i$ now have equal sums. If $x_k-y_l$ is not included in this sum then these subsets are subsets of the original sets and we are finished. If $x_k-y_l$ is included in this sum then replace it by $x_k$ and re-introduce $y_l$ to obtain proper subsets for the original sets of numbers.