Color a 1000 by 1000 grid with 0s and 1s Show that one could remove 990 rows with a 1 remaining in every column, OR delete 990 columns with 0 in remaining in every row.
My approach: let $r_i$ be the number of 1s in row i and $c_i$ be the number of 0s in column i but then I am stuck. Any hint would be appreciated!
 A: Suppose it is impossible to delete 990 rows so that there is a 1 in every column. This means, any 10 rows always have a column with only 0s. For rows $10i ... 10i + 9$, denote the all-zero column by $z_i$, for $i = 0...9$. This gives 10 columns. 
If we now delete all but these 10 columns $z_i$, we guaranteed a 0 in every row. (So column $z_0$ will have 0 in the first 10 rows, column $z_1$ will have a 0 in the next 10 rows, and so on.)

This is a variation on the following observation:
In any rectangle colored with 0s and 1s, we have both the following being true:


*

*either a 1 in every column, or a 0 in every row.

*either a 1 in every row, or a 0 in every column.


(To see that this is true, take the first statement as an example. If we do NOT have a one in every column, there is a column with all 0s, and so we are guaranteed a 0 in every row.)
A: Not a solution, too long for comments. I was thinking on the problem and decided to write my thoughts; my first idea was to use the pideonhole principle and it would be straightforward; it is more complicated than I thought.
A second idea is to try to use some linear indepedence argument, considering rows and columns as vectors on $\mathbb{Z}_2^{1000}$. This sounds familiar, but I could not find a way to express the conditions on this new angle.
Any comments/suggestions are welcome.

Pidgeonhole attempt.
Let us fix some notation. Denote $[n] = \{1, 2, \ldots, n\}$ and also for some set $X$ let
$$
\binom{X}{n} = \{Y \subset X; |Y| = n\}
$$
the family of all subsets of $X$ with precisely $n$ elements. Using your notation for $r_i$ and $c_i$, we have
$$
\sum_1^{1000} c_i + \sum_1^{1000} r_i = 1000 \cdot 1000 = 10^6,
$$
since we are counting each $1$ and each $0$ exactly once. For some 
$X \in \binom{[1000]}{10}$, let us denote
$$
r_X = \sum_{i \in X} r_i.
$$
Using the pidgeonhole principle, if you have a grid that is $10 \times 1000$ that has at least $999 \cdot 10 + 1$ elements being $1$, then you found a selection of rows that satisfies the first statement. Otherwise, for every $X \in \binom{[1000]}{10}$ you have
$$
r_X \le 9990.
$$
Now, we make a counting in two ways argument. Notice that
$$
\sum_{X \in \binom{[1000]}{10}} r_X = \binom{999}{9} \cdot \sum_{1}^{1000} r_i,
$$
since every row belongs to precisely $\binom{999}{9}$ sets of $10$ rows (just choose $9$ other rows from the remaining $999$). The computations above show that if we do not have the first statement then
$$
\sum_{1}^{1000} r_i
\le \frac{\binom{1000}{10}}{\binom{999}{9}} \cdot 9990
= \frac{1000!}{999!} \cdot \frac{9! \cdot 990!}{10! \cdot 990!} \cdot 9990
= 1000 \cdot 999
$$
and thus the average row has at most $999$ ones.
Symmetric argument for $c_i$. Analogously, if we have some $1000 \times 10$ grid with at least $999 \cdot 10 + 1$ entries being zero then every row must have a zero. The same double counting argument proves that
$$
\sum_1^{1000} c_i \le 1000 \cdot 999.
$$
