Let $n$ be a positive integer. Let $S$ be a subset of points on the plane with these conditions: 

$i)$  There does not exist $n$ lines in the plane such that every element of $S$ be on at least one of them.  

$ii)$ for all $X \in S$ there exists $n$ lines in the plane such that every element of $S - {X} $ be on at least one of them.

Find maximum of $\mid S\mid$.

[i]Proposed by Erfan Salavati[/i]
1. Let \( S \) be a subset of points on the plane such that there does not exist \( n \) lines in the plane such that every element of \( S \) is on at least one of them. This is condition \( i) \).
2. For each point \( X \in S \), there exists \( n \) lines in the plane such that every element of \( S - \{X\} \) is on at least one of them. This is condition \( ii) \).
3. Let \( (x_i, y_i) \) for \( 1 \le i \le |S| \) be the points in \( S \). For a point \( X_i = (x_i, y_i) \), consider the \( n \) lines that cover \( S - \{X_i\} \), say \( a_j x + b_j y + c_j = 0 \) for \( 1 \le j \le n \).
4. By condition \( i) \), \( X_i \) is not on any of these \( n \) lines, so the polynomial 
   \[
   f_i(x, y) = \prod_{j=1}^{n} (a_j x + b_j y + c_j)
   \]
   is nonzero for \( X_i \) and is zero for any other point in \( S \).
5. We have that \( f_i \in \mathbb{R}[x, y] \) and has degree at most \( n \). The 2-variable polynomials of degree at most \( n \) form a vector space over the real numbers, and a basis has cardinality 
   \[
   \frac{(n+1)(n+2)}{2}
   \]
   (take the standard basis \( \mathcal{B} = \{ x^k y^\ell \mid k + \ell \le n \} \)).
6. It is easy to see that the polynomials \( f_i \) are linearly independent over the reals. Indeed, if 
   \[
   \sum_{i=1}^{|S|} \alpha_i f_i(x, y) = 0,
   \]
   by plugging \( (x_i, y_i) \) into the previous relation, all polynomials cancel with the exception of \( f_i \), hence \( \alpha_i = 0 \). Therefore, \( |S| \le \frac{(n+1)(n+2)}{2} \).
7. An example can be given using the aforementioned triangular array.

The final answer is \(\boxed{\frac{(n+1)(n+2)}{2}}\).