An $m\times n$ table has $0$ or $1$ in each cell. ($m$, $n$ even; at least one $1$.) Show that there exists a "cross" of cells whose sum is odd. A (hard) combinatorics problem:

There is a $m \times n$ table, and you write $0$ or $1$ in every cell of the table. Show that there exists a "cross" such that the sum of integers written in the cross is odd. (There is at least one cell that has $1$ written on it. $n$ and $m$ are even.)
A "cross" is the "+"-like shape that is made when a row and a column intersects.

My try:
I assumed that all the crosses are even and tried to show a contradiction, but I could not progress further.
 A: *

*notations

$ d ( i , j ) $ : The number written in the $ i $ -th row , $ j $ -th column.
$ a _ i $ : $ \sum _{ j = 1 } ^ { n } { d( i , j ) } $
$ b _ j $ : $ \sum _{ i = 1 } ^ { m } { d( i , j ) } $
So, $ a_ i $ is the summation of numbers written in the $ i$ -th row, and $ b_ j$ is the summation of numbers written in the $j$ -th column.
$ S $ : $ \sum _{ 1 \le i \le m , 1 \le j \le n } { d( i , j ) } $
So, $ S $ is the summation of all the numbers in the table.
Note that $ S = \sum _{ i = 1 } ^{ m } { a_ i } = \sum _{ j= 1 } ^{ n } { b_ j } $ since the summation of all the numbers written in the table is equal to the the summation of the sum of numbers written in the $ i $ -th row , and in the same way, is equal to the summation of columns.
$ c ( i , j ) $ : The summation of the numbers written in the cross made by the $ i $ -th row and the $ j $ -th column.
Note that $ c( i , j ) = a_i + b_j - d ( i , j ) $ by the definition of $ c (i , j ) $.
Let us assume that the summation of the numbers written in the cross is always even for the sake of contradiction. So, we are assuming that $ c( i , j ) $ is even for all $ 1 \le i \le m , 1 \le j \le n $ .
For a fixed integer i and an integer j ( $ 1 \le j \le n $ ), think of the cross made by the $ i $ -th row and the $ j $ -th column.
The summation of these crosses are
\begin{align*}
 \sum _{ j = 1 } ^{ n } c( i , j ) &= \sum _{ j = 1 } ^{ n } ( a_ i + b_j - d(i , j ) ) \\
&= \sum _{ j = 1 } ^{ n } { b_j } + na_i - \sum _{ j = 1 } ^{ n } { d( i , j) } \\
&= \sum _{ j = 1 } ^{ n } { b_ j } + ( n - 1 )a_ i \\
&= S + ( n - 1 )a_ i .
\end{align*}
Since we assumed that $ c( i , j ) $ is even for all $ 1 \le i \le m , 1 \le j \le n $, we can say that
$S + ( n - 1 )a_ i \equiv 0 \pmod 2 $
Since $n \equiv 0 \pmod 2$, $ S + a_i \equiv 0 \pmod 2 $.
Therefore, we can say that $ a_i \equiv S \pmod 2 $ for a fixed integer $i$.
If we repeat the process for all $ i$, we get $ a_i \equiv S \pmod 2 $ for all $ i $.
In the same way, we can get $ b_j \equiv S \pmod 2$ for all $j$ .
So, for all $i, j $, $ a_i \equiv b_j \pmod 2 $.
Meanwhile, $ c( i, j) = a_i + b_j -d( i , j ) \equiv 0 \pmod 2 $
So, $ d( i, j) \equiv 0 \pmod 2 $ for all $ i,j $.
Note that $ d( i, j) \in \{ 0,1 \} $, so we get that $ d( i, j) =0 $ for all $ i,j $ , which is a contradiction.
So, there exists a pair of positive integers $ ( i , j ) $ such that $ c(i, j) \equiv 1 \pmod 2 $.
