A good approximation of $\pi$ is $3.14.$ Find the least positive integer $d$ such that if the area of a circle with diameter $d$ is calculated using the approximation $3.14,$ the error will exceed $1.$
1. We start by noting that the area \( A \) of a circle with radius \( r \) is given by:
   \[
   A = \pi r^2
   \]
   If we use the approximation \( \pi \approx 3.14 \), the approximate area \( A' \) is:
   \[
   A' = 3.14 r^2
   \]

2. The error \( E \) in the area calculation using the approximation is:
   \[
   E = \left| \pi r^2 - 3.14 r^2 \right| = \left| (\pi - 3.14) r^2 \right|
   \]

3. Given that \( 3.14159 < \pi < 3.14160 \), we can bound the error:
   \[
   0.00159 r^2 < E < 0.00160 r^2
   \]

4. We need to find the smallest positive integer \( d \) such that the error \( E \) exceeds 1. Therefore, we set up the inequality:
   \[
   0.00159 r^2 > 1 \quad \text{or} \quad 0.00160 r^2 > 1
   \]

5. Solving for \( r^2 \) in both inequalities:
   \[
   r^2 > \frac{1}{0.00159} \approx 628.93
   \]
   \[
   r^2 > \frac{1}{0.00160} = 625
   \]

6. Taking the square root of both sides to find \( r \):
   \[
   r > \sqrt{628.93} \approx 25.1
   \]
   \[
   r > \sqrt{625} = 25
   \]

7. Since \( r \) must be an integer or half-integer, we test \( r = 25 \) and \( r = 25.5 \).

8. For \( r = 25 \):
   \[
   \text{Actual area} = \pi \cdot 25^2 \approx 3.14159 \cdot 625 \approx 1963.49375 \quad \text{to} \quad 3.14160 \cdot 625 \approx 1963.5
   \]
   \[
   \text{Approximate area} = 3.14 \cdot 25^2 = 3.14 \cdot 625 = 1962.5
   \]
   \[
   \text{Error} = 1963.49375 - 1962.5 \approx 0.99375 \quad \text{to} \quad 1963.5 - 1962.5 = 1
   \]
   The error is less than or equal to 1.

9. For \( r = 25.5 \):
   \[
   \text{Actual area} = \pi \cdot 25.5^2 \approx 3.14159 \cdot 650.25 \approx 2042.8189 \quad \text{to} \quad 3.14160 \cdot 650.25 \approx 2042.8254
   \]
   \[
   \text{Approximate area} = 3.14 \cdot 25.5^2 = 3.14 \cdot 650.25 = 2041.785
   \]
   \[
   \text{Error} = 2042.8189 - 2041.785 \approx 1.0339 \quad \text{to} \quad 2042.8254 - 2041.785 = 1.0404
   \]
   The error exceeds 1.

10. Since the error exceeds 1 for \( r = 25.5 \), the corresponding diameter \( d \) is:
    \[
    d = 2r = 2 \times 25.5 = 51
    \]

The final answer is \( \boxed{51} \).