Alice and Bob play a game together as a team on a $100 \times 100$ board with all unit squares initially white. Alice sets up the game by coloring exactly $k$ of the unit squares red at the beginning. After that, a legal move for Bob is to choose a row or column with at least $10$ red squares and color all of the remaining squares in it red. What is the
smallest $k$ such that Alice can set up a game in such a way that Bob can color the entire board red after finitely many moves?

Proposed by [i]Nikola Velov, Macedonia[/i]
1. **First Claim: Alice can color 100 squares and Bob can color the entire board.**

   Let's name the rows \( A_1, A_2, A_3, \ldots, A_{100} \) and the columns \( B_1, B_2, B_3, \ldots, B_{100} \). If Alice colors the squares \( A_1B_1, A_1B_2, \ldots, A_1B_{10}, A_2B_1, A_2B_2, \ldots, A_2B_{10}, \ldots, A_{10}B_{10} \), then Bob can color the entire board.

   - Alice colors the first 10 rows and the first 10 columns, making a \(10 \times 10\) grid of red squares.
   - Each row and each column in this \(10 \times 10\) grid has exactly 10 red squares.
   - Bob can then choose any row or column with at least 10 red squares and color the entire row or column red.
   - This process can be repeated until the entire board is red.

2. **Second Claim: Alice has to color at least 100 squares.**

   Proof: Suppose the number of squares Alice colors is less than 100.

   - Without loss of generality (WLOG), suppose Bob starts by coloring rows. Initially, he colors row \( A_1 \), which means there are at least \( 10A_1 \) red squares.
   - If \( A_1 < 10 \), he then colors column \( B_1 \), which means there are at least \( B_1(10 - A_1) \) more red squares.
   - If \( B_1 < 10 \), he colors row \( A_2 \), adding \( A_2(10 - B_1) \) more red squares.
   - This process continues, alternating between rows and columns, until either \( A_1 + A_2 + \ldots + A_k = 10 \) or \( B_1 + B_2 + \ldots + B_t = 10 \).

   At the \( k \)-th step, if either \( A_1 + \ldots + A_k = 10 \) or \( B_1 + \ldots + B_t = 10 \), Bob finishes coloring.

   - The total number of red squares is given by:
     \[
     10(A_1 + \ldots + A_k + B_1 + \ldots + B_t) - (A_1 + \ldots + A_k)(B_1 + \ldots + B_t)
     \]
   - Since \( A_1 + \ldots + A_k = 10 \) or \( B_1 + \ldots + B_t = 10 \), we have:
     \[
     10(A_1 + \ldots + A_k + B_1 + \ldots + B_t) - (A_1 + \ldots + A_k)(B_1 + \ldots + B_t) < 100
     \]
   - This implies:
     \[
     100 < 100
     \]
   - This is a contradiction, so Alice must color at least 100 squares.

Therefore, the smallest \( k \) such that Alice can set up the game in such a way that Bob can color the entire board red after finitely many moves is \( k = 100 \).

The final answer is \( \boxed{100} \).