Problem 1518

1. Define the labels and cumulative sums:
Let $r_1, r_2, \ldots, r_N$ be the labels of the red cards and $b_1, b_2, \ldots, b_N$ be the labels of the blue cards. Define the cumulative sums:
$$x_i = \sum_{j=1}^i r_j \quad \text{for} \quad 1 \le i \le N$$
$$y_i = \sum_{j=1}^i b_j \quad \text{for} \quad 1 \le i \le N$$

2. **Construct the set of points $S$:**
Consider the set of points $S = \{0, 1, 2, \ldots, N\} \times \{0, 1, 2, \ldots, N\}$. For each point $(a, b) \in S$, label it with $x_a + y_b$.

3. Apply the Pigeonhole Principle:
We claim that some two points in $S$ have the same label. To show this, we will use the Pigeonhole Principle.

4. **Define the sets $S_i$:**
For each $0 \le i \le N$, let $S_i$ be the set of points $(a, b) \in S$ such that $\max(|a-N|, |b|) = i$. This means $S_i$ contains points on the boundary of the square grid at distance $i$ from the point $(N, 0)$.

5. Label the points and apply discrete continuity:
Let $\ell$ be the label of the point $(N, 0)$, which is $x_N + y_0 = x_N$. For every $0 \le i \le N$, note that since $(N-i, 0)$ and $(N, i)$ are both in $S_i$, by "discrete continuity" there must exist some $(a, b) \in S_i$ such that the label of $(a, b)$ is in the interval $[\ell, \ell + N)$.

6. Count the distinct labels:
Since the $S_i$'s are mutually disjoint and there are $N+1$ of them, we now know the existence of $N+1$ distinct points in $S$ which all have labels in $[\ell, \ell+N)$. By the Pigeonhole Principle, there must be at least two points in $S$ with the same label.

7. Identify the points with the same label:
Hence, we can pick $(a, b), (c, d) \in S$ such that $a < c$ and $d < b$ and $(a, b)$ and $(c, d)$ have the same label.

8. Derive the subset sums:
By the definition of our table, we have:
$$x_c + y_d = x_a + y_b$$
Rearranging, we get:
$$x_c - x_a = y_b - y_d$$
This implies that the sum of the numbers on the chosen red cards (from $a+1$ to $c$) equals the sum of the numbers on the chosen blue cards (from $d+1$ to $b$).

Therefore, we have shown that it is possible to choose a (non-empty) subset of the red cards and a (non-empty) subset of the blue cards such that the sum of the numbers on the chosen red cards equals the sum of the numbers on the chosen blue cards.

$\blacksquare$