Problem 1375

1. Define the sets and the board:
Let $C = \{C_1, C_2, \ldots, C_{N+1}\}$ be the set of columns of the given board, where each column $C_j = (x_{1j}, x_{2j}, \ldots, x_{Nj})$ and $x_{ij} \in \mathbb{Z}$ for all $1 \le i \le N$ and $1 \le j \le N+1$.

2. Define the power set and related sets:
Let $\mathbb{P}(C)$ be the power set of $C$, i.e., the set of all subsets of $C$, including the empty set and $C$ itself. Then $\big|\mathbb{P}(C)\big| = 2^{N+1}$.

Let $\mathbb{A} := \mathbb{P}(C) - \{\emptyset\}$. Then $\big|\mathbb{A}\big| = 2^{N+1} - 1$.

Let $\mathbb{B} := \{(y_1, y_2, \ldots, y_N) : y_i \in \{0,1\}, 1 \le i \le N\}$. Then $\big|\mathbb{B}\big| = 2^N$.

3. **Pairing elements of $\mathbb{A}$ with $\mathbb{B}$:**
We pair the elements of $\mathbb{A}$ with the elements of $\mathbb{B}$ according to the following rule:
$$\{C_{j_1}, C_{j_2}, \ldots, C_{j_k}\} \longrightarrow (z_1, z_2, \ldots, z_N)$$
if
$$x_{ij_1} + x_{ij_2} + \ldots + x_{ij_k} \equiv z_i \pmod{2}$$
for all $1 \le i \le N$ and $1 \le k \le N+1$.

4. Application of the pigeonhole principle:
Since $2^{N+1} - 1 > 2^N$ for all $N \in \mathbb{Z}^+$, by the pigeonhole principle, there must be at least two different elements $X, Y \in \mathbb{A}$ that are paired with the same element in $\mathbb{B}$.

5. Constructing the desired set of columns:
Consider the sets $X$ and $Y$ which are paired with the same element in $\mathbb{B}$. Define the symmetric difference of $X$ and $Y$ as:
$$(X \setminus Y) \cup (Y \setminus X)$$
Since $X \neq Y$, this set is non-empty.

6. Ensuring the sum in each row is even:
If we remove all columns except the ones belonging to the set $(X \setminus Y) \cup (Y \setminus X)$, the sum of the numbers left uncrossed out in each row will be even. This is because the sums corresponding to $X$ and $Y$ are the same modulo 2, and the symmetric difference will ensure that the sums in each row are even.

$\blacksquare$