If all vertices of a triangle on the square grid are grid points, then the triangle is called a [i]lattice[/i] triangle. What is the area of the lattice triangle with (one) of the smallest area, if one grid  has area $1$ square unit?
1. **Understanding the Problem:**
   We need to find the area of the smallest lattice triangle on a square grid where each grid square has an area of 1 square unit.

2. **Using Pick's Theorem:**
   Pick's Theorem states that for a lattice polygon, the area \( A \) can be calculated using the formula:
   \[
   A = I + \frac{B}{2} - 1
   \]
   where \( I \) is the number of interior lattice points and \( B \) is the number of boundary lattice points.

3. **Setting Up the Smallest Lattice Triangle:**
   To find the smallest lattice triangle, we consider a triangle with no interior points (\( I = 0 \)) and the minimum number of boundary points. A triangle must have at least 3 boundary points (one at each vertex).

4. **Calculating the Area:**
   - Let \( I = 0 \) (no interior points).
   - Let \( B = 3 \) (one point at each vertex of the triangle).

   Substituting these values into Pick's Theorem:
   \[
   A = 0 + \frac{3}{2} - 1 = \frac{3}{2} - 1 = \frac{1}{2}
   \]

5. **Verification:**
   To verify, consider a right triangle with vertices at \((0,0)\), \((1,0)\), and \((0,1)\). This triangle has:
   - 3 boundary points (the vertices).
   - 0 interior points.

   Using the formula for the area of a right triangle:
   \[
   A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}
   \]

   This confirms that the area calculated using Pick's Theorem is correct.

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