Determine the smallest positive constant $k$ such that no matter what $3$ lattice points we choose the following inequality holds: 
$$ L_{\max} - L_{\min} \ge \frac{1}{\sqrt{k} \cdot L_{max}} $$ 
where $L_{\max}$, $L_{\min}$ is the maximal and minimal distance between chosen points. 
1. **Observations:**
   - We observe that \( L_{\max} \) is strictly greater than \( L_{\min} \), as the lattice points create a triangle with rational area. However, the area of an equilateral triangle is irrational.
   - Both \( L_{\max} \) and \( L_{\min} \) are square roots of some positive integers. This follows due to the fact they can be expressed as \( \sqrt{x^2 + y^2} \), where \( x, y \) are their respective lengths of projections on the coordinate axis.
   - By examining small triangles, we can get a lower bound \( k \ge 4 \), which is achieved in the triangle \( A(0,0), B(0,2), C(2,1) \). It is easy to verify this.

2. **Verification for \( k = 4 \):**
   - Rewrite the inequality as:
     \[
     L_{\max}^2 - L_{\max} L_{\min} \ge \frac{1}{\sqrt{k}}
     \]
   - Denote \( L_{\max} = \sqrt{a} \) and \( L_{\min} = \sqrt{a - b} \), where \( b < a \) and \( a, b \in \mathbb{N} \).
   - The inequality becomes:
     \[
     a - \sqrt{a(a - b)} \ge \frac{1}{2}
     \]
   - We need to show that:
     \[
     a - \sqrt{a(a - b)} \ge a - \sqrt{a(a - 1)}
     \]
   - Simplify the inequality:
     \[
     a - \sqrt{a(a - 1)} \ge \frac{1}{2}
     \]
   - This can be rewritten as:
     \[
     \left( a - \frac{1}{2} \right)^2 \ge a(a - 1)
     \]
   - Expanding and simplifying:
     \[
     a^2 - a + \frac{1}{4} \ge a^2 - a
     \]
   - This inequality is obviously true since:
     \[
     \frac{1}{4} \ge 0
     \]

Thus, the smallest positive constant \( k \) such that the given inequality holds for any three lattice points is \( k = 4 \).

The final answer is \( \boxed{4} \).