A positive integer $n$ is fixed. Numbers $0$ and $1$ are placed in all cells (exactly one number in any cell) of a $k \times n$ table ($k$ is a number of the rows in the table, $n$ is the number of the columns in it). We call a table nice if the following property is fulfilled: for any partition of the set of the rows of the table into two nonempty subsets $R$[size=75]1[/size] and $R$[size=75]2[/size] there exists a nonempty set $S$ of the columns such that on the intersection of any row from $R$[size=75]1[/size] with the columns from $S$ there are even number of $1's$ while on the intersection of any row from $R$[size=75]2[/size] with the columns from $S$ there are odd number of $1's$.
 Find the greatest number of $k$ such that there exists at least one nice $k \times n$ table.
1. **Define the problem in terms of matrices:**
   Let \( M \in \mathbb{Z}_2^{k \times n} \) be the matrix representing the table, where \( m_{i,j} \) is the entry in the \( i \)-th row and \( j \)-th column, and the size of \( M \) is \( k \times n \).

2. **Formulate the problem using linear algebra:**
   The problem requires that for any partition of the rows into two nonempty subsets \( R_1 \) and \( R_2 \), there exists a nonempty set \( S \) of columns such that:
   - For any row in \( R_1 \), the sum of the entries in the columns of \( S \) is even.
   - For any row in \( R_2 \), the sum of the entries in the columns of \( S \) is odd.

3. **Translate the condition into a system of linear equations:**
   If \( k > n \), then for any vector \( r \in \mathbb{Z}_2^k \), there exists a vector \( s \in \mathbb{Z}_2^n \) such that \( Ms = r \). This is because the number of possible \( r \) vectors is \( 2^k \), and the number of possible \( s \) vectors is \( 2^n \). If \( k > n \), then \( 2^k > 2^n \), which means there are more possible \( r \) vectors than \( s \) vectors, making it impossible to satisfy the condition for all \( r \).

4. **Conclude that \( k \leq n \):**
   Since \( k > n \) leads to a contradiction, we conclude that \( k \) must be at most \( n \).

5. **Verify if \( k = n \) can be a solution:**
   If \( k = n \), consider the identity matrix \( M = I \) (where the table has ones on its main diagonal and zeros elsewhere). This matrix is "nice" because:
   - For any partition of the rows, we can choose \( S \) to be the set containing the column corresponding to the row in \( R_2 \). This ensures that the sum of the entries in \( S \) for rows in \( R_1 \) is zero (even), and for rows in \( R_2 \) is one (odd).

6. **Conclude the greatest number of \( k \):**
   Therefore, the greatest number of \( k \) such that there exists at least one nice \( k \times n \) table is \( k = n \).

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