Problem 1348

1. **Step 1: Analyze the given polynomial for specific values of $n$**
- Let $n = 0$:
$$P(0) = d$$
Since $P(0)$ is an integer, $d$ must be an integer.

- Let $n = 1$:
$$P(1) = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d$$
Since $d$ is an integer and $P(1)$ is an integer, $a + b + c$ must also be an integer.

- Let $n = 2$:
$$P(2) = a(2)^3 + b(2)^2 + c(2) + d = 8a + 4b + 2c + d$$
Since $d$ is an integer and $P(2)$ is an integer, $8a + 4b + 2c$ must also be an integer.

2. **Step 2: Derive conditions from the polynomial for $n = 2$**
- Consider the difference:
$$(8a + 4b + 2c + d) - 2(a + b + c + d) = 8a + 4b + 2c - 2a - 2b - 2c = 6a + 2b$$
Since $8a + 4b + 2c + d$ and $a + b + c + d$ are integers, $6a + 2b$ must be an integer.

3. **Step 3: Derive conditions from the polynomial for $n = 3$**
- Let $n = 3$:
$$P(3) = a(3)^3 + b(3)^2 + c(3) + d = 27a + 9b + 3c + d$$
Since $d$ is an integer and $P(3)$ is an integer, $27a + 9b + 3c$ must also be an integer.

- Consider the difference:
$$(27a + 9b + 3c + d) - 3(a + b + c + d) = 27a + 9b + 3c - 3a - 3b - 3c = 24a + 6b$$
Since $27a + 9b + 3c + d$ and $a + b + c + d$ are integers, $24a + 6b$ must be an integer.

4. Step 4: Combine the conditions
- From $6a + 2b$ being an integer and $24a + 6b$ being an integer, we can derive:
$$4(6a + 2b) - (24a + 6b) = 24a + 8b - 24a - 6b = 2b$$
Since $2b$ is an integer, $b$ must be an integer.

- From $6a + 2b$ being an integer and $b$ being an integer:
$$6a + 2b - 2b = 6a$$
Since $6a$ is an integer, $a$ must be a rational number such that $6a$ is an integer. Therefore, $a$ must be of the form $\frac{k}{6}$ where $k$ is an integer.

5. **Step 5: Determine the minimal value of $a$**
- Since $a > 0$, the smallest positive value for $a$ is when $k = 1$:
$$a = \frac{1}{6}$$

Conclusion:
The minimal value of $a$ is $\boxed{\frac{1}{6}}$.