All-Soviet-Union Mathematical Olympiad

Solution:

We prove the general case. Let the first sequence be $a_1, a_2, \ldots, a_m$ and the second sequence be $b_1, b_2, \ldots, b_n$, where $0 < a_i \leq n$ and $0 < b_j \leq m$. Put $s_k = a_1 + a_2 + \ldots + a_k$, $t_k = b_1 + b_2 + \ldots + b_k$. Assume $s_m > t_n$ (if they are equal, then we are done).

Let $f(i)$ be the smallest $k$ such that $s_k \geq t_i$. If it is equal, we are done, so assume $s_k > t_i$. Now consider the $n$ numbers $s_{f(i)} - t_i$. Each is at least $1$ and at most $n-1$ (if it was $n$ or more then $s_{f(i)-1} \geq s_{f(i)} - n \geq t_i$, contradicting the minimality of $f(i)$). So there must be two the same. So we have $s_{f(i)} - t_i = s_{f(i)} - t_j$ for some $i > j$ and hence $a_{f(j)+1} + a_{f(j)+2} + \ldots + a_{f(i)} = b_{j+1} + b_{j+2} + \ldots + b_i$.