Find two subsets with a common sum in two sequences For a positive integer $n$ and two integer sequences $a_1,a_2...a_n$ and $b_1,b_2...b_n$ where $\forall i$, $a_i,b_i \in [1,n]$, I want to find two non-empty subsets, one in each sequence, with the same sum.
i.e. I want to find $1 \leq i_1<i_2 <\cdots <i_p \leq n$ and $1 \leq j_1<j_2 <\cdots < j_q \leq n$ where $p,q \geq 1$ and $\sum_{x=1}^p a_{i_x}=\sum_{y=1}^q b_{j_y}$.
From trying small cases, I highly suspect that such subsets exist for any $a$ and $b$, but I can't find a rigorous proof for that.
 A: This solution is paraphrased from Mathematical Mind-Benders by Peter Winkler, a math puzzle anthology where this puzzle appears. 

For each $k\ge 0$, let 
$$
s_k=a_1+a_2+\dots+a_k,\\
t_k=b_1+b_2+\dots+b_k,
$$
with the convention that $s_0=t_0=0$. Assume WLOG that $s_n\le t_n$, which can achieved by possibly switching $a$ and $b$. Further assume $s_n<t_n$, since in case $s_n=t_n$ the entire sets work. 
For each $k=0,1,\dots,n$, let $k'$ be the largest element of $\{0,1,\dots,n-1\}$ for which $s_{k}\ge t_{k'}$. Note that $k'$ exists; there is at least one index $k'$ for which $s_k\ge t_{k'}$, since $s_k\ge t_0$. 
We then have that
$$
0\le s_k-t_{k'}\le n-1.
$$
The left inequality is obvious. The right inequality follows by the maximality of $k'$; if $s_k\ge t_{k'}+n$, then you would also have $s_k\ge t_{k'+1}$. 
Since $k$ can take any of $n+1$ values between $0$ and $n$ inclusive, but $s_{k}-t_{k'}$ can only take $n$ values between $0$ and $n-1$ inclusive, by the pigeonhole principle, there must exist indices $k\ge h$ for which
$$
s_k-t_{k'}=s_h-t_{h'},
$$
which implies
$$
s_k-s_h=a_{h+1}+\dots+a_k=b_{h'+1}+\dots+b_{k'}=t_{k'}-t_{h'},
$$
so we have found our two equal subsets.
