In a checked $ 17\times 17$ table, $ n$ squares are colored in black. We call a line any of rows, columns, or any of two diagonals of the table. In one step, if at least $ 6$ of the squares in some line are black, then one can paint all the squares of this line in black. 
Find the minimal value of $ n$ such that for some initial arrangement of $ n$ black squares one can paint all squares of the table in black in some steps.
To find the minimal value of \( n \) such that for some initial arrangement of \( n \) black squares one can paint all squares of the \( 17 \times 17 \) table in black in some steps, we need to consider the conditions under which a line (row, column, or diagonal) can be painted black. Specifically, a line can be painted black if it contains at least 6 black squares.

1. **Assume \( n \leq 26 \) and derive a contradiction:**
   - Initially, there are \( 17^2 - n = 289 - n \) white squares.
   - If \( n \leq 26 \), then there are at least \( 289 - 26 = 263 \) white squares.
   - In each step, at most \( 17 - 6 = 11 \) new squares can be painted black.
   - Therefore, at least \( \left\lceil \frac{263}{11} \right\rceil = 24 \) steps are required to paint all squares black.

2. **Define sets and variables:**
   - Let \( B \) be the set of initially black squares.
   - Let \( A_i \) be the line chosen in the \( i \)-th step to paint.
   - Let \( B_i = A_i \cap B - \left( \bigcup_{k=1}^{i-1} A_k \right) \).
   - Let \( b_i = |B_i| \).

3. **Analyze the intersections:**
   - For any two lines \( A_i \) and \( A_j \), \( |A_i \cap A_j| \leq 1 \) for \( i \neq j \).
   - If \( A_i \) and \( A_j \) are both rows or both columns, \( A_i \cap A_j = \emptyset \).

4. **Calculate the minimum number of initially black squares:**
   - Since there must be at least 6 black squares in a line to paint it, we have:
     \[
     b_i \geq 6 - \sum_{j < i} |A_i \cap A_j|
     \]
   - If \( A_i \) is a row:
     \[
     b_i \geq 6 - |\{j | j < i, A_j \text{ is column-oriented}\}|
     \]
   - If \( A_i \) is a column:
     \[
     b_i \geq 6 - |\{j | j < i, A_j \text{ is row-oriented}\}|
     \]
   - If \( A_i \) is a diagonal:
     \[
     b_i \geq 6 - (i - 1)
     \]

5. **Count the number of steps:**
   - Since there are at least 24 steps, at least 6 of the sets \( A_i \) must be row-oriented or column-oriented.
   - Let \( i \) be the smallest positive integer such that at least 6 of the sets \( A_1, A_2, \ldots, A_i \) are all row-oriented or all column-oriented.
   - Without loss of generality, suppose the latter is true.
   - Let \( r \) be the number of rows, \( c \) be the number of columns, and \( d \) be the number of diagonals among the sets \( A_1, A_2, \ldots, A_i \).
   - Note \( r + d = 6 \) and \( d \leq 2 \).

6. **Calculate the total number of initially black squares:**
   - We have:
     \[
     |B| \geq \sum_{k=1}^{i} b_k \geq 6i - \left( \sum_{k | A_k \text{ row}} |\{j | j < k, A_j \text{ row-oriented}\}| + \sum_{k | A_k \text{ column}} |\{j | j < k, A_j \text{ column-oriented}\}| + \sum_{k | A_k \text{ diagonal}} (k - 1) \right)
     \]
   - The first sum counts the number of pairs \( (j, k) \) such that \( 1 \leq j < k \leq i \) and \( A_j \) is a row, \( A_k \) is a column.
   - The second sum counts the number of pairs \( (j, k) \) such that \( 1 \leq j < k \leq i \) and \( A_j \) is a column, \( A_k \) is a row.
   - The third sum counts the number of pairs \( (j, k) \) such that \( 1 \leq j < k \leq i \) and \( A_k \) is a diagonal.

7. **Simplify the expression:**
   - Using \( c = i - r - d \) and \( r + d = 6 \):
     \[
     |B| \geq 6i - (r(i - r - d) + d(i - d) + \binom{d}{2})
     \]
   - Simplifying further:
     \[
     |B| \geq (6 - r - d)i + r^2 + rd + d^2 - \binom{d}{2}
     \]
   - The minimum value is attained when \( (r, d) = (4, 2) \):
     \[
     |B| \geq 4^2 + 4 \cdot 2 + 2^2 - \binom{2}{2} = 16 + 8 + 4 - 1 = 27
     \]

8. **Construct an arrangement with 27 black squares:**
   - Number the rows and columns 1-17.
   - Color the following squares black:
     \[
     (7,1), (8,1), (9,1), (10,1), (7,2), (8,2), (9,2), (10,2), (7,16), (8,16), (9,16), (10,16), (7,17), (8,17), (9,17), (10,17), (7,7), (8,8), (9,9), (10,10), (11,11), (12,12), (11,7), (10,8), (8,10), (7,11), (6,12)
     \]
   - To paint all squares black, first paint both diagonals, then columns 1, 2, 16, 17, then rows 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, and finally all columns that still have at least 1 white square.

The final answer is \( \boxed{ 27 } \).