Let $P(x)=x^3+ax^2+bx+c$ where $a,b,c$ are positive real numbers. If $P(1)\geq 2$ and $P(3)\leq 31$, how many of integers can $P(4)$ take?

$ 
\textbf{(A)}\ 3
\qquad\textbf{(B)}\ 4
\qquad\textbf{(C)}\ 5
\qquad\textbf{(D)}\ 6
\qquad\textbf{(E)}\ \text{None of above}
$
1. Given the polynomial \( P(x) = x^3 + ax^2 + bx + c \) where \( a, b, c \) are positive real numbers, we need to find the possible integer values of \( P(4) \) given the constraints \( P(1) \geq 2 \) and \( P(3) \leq 31 \).

2. First, evaluate \( P(1) \):
   \[
   P(1) = 1^3 + a \cdot 1^2 + b \cdot 1 + c = 1 + a + b + c
   \]
   Given \( P(1) \geq 2 \), we have:
   \[
   1 + a + b + c \geq 2 \implies a + b + c \geq 1
   \]

3. Next, evaluate \( P(3) \):
   \[
   P(3) = 3^3 + a \cdot 3^2 + b \cdot 3 + c = 27 + 9a + 3b + c
   \]
   Given \( P(3) \leq 31 \), we have:
   \[
   27 + 9a + 3b + c \leq 31 \implies 9a + 3b + c \leq 4
   \]

4. Now, evaluate \( P(4) \):
   \[
   P(4) = 4^3 + a \cdot 4^2 + b \cdot 4 + c = 64 + 16a + 4b + c
   \]

5. We need to find the range of \( 16a + 4b + c \) given the constraints:
   \[
   a + b + c \geq 1 \quad \text{and} \quad 9a + 3b + c \leq 4
   \]

6. From the second inequality, solve for \( c \):
   \[
   c \leq 4 - 9a - 3b
   \]

7. Substitute \( c \) from the first inequality:
   \[
   a + b + (4 - 9a - 3b) \geq 1 \implies 4 - 8a - 2b \geq 1 \implies 3 \geq 8a + 2b \implies \frac{3}{2} \geq 4a + b
   \]

8. Now, substitute \( c \) in \( P(4) \):
   \[
   P(4) = 64 + 16a + 4b + (4 - 9a - 3b) = 68 + 7a + b
   \]

9. Given \( 4a + b \leq \frac{3}{2} \), we have:
   \[
   7a + b \leq 7a + \frac{3}{2} - 4a = 3a + \frac{3}{2}
   \]

10. Therefore, the range of \( P(4) \) is:
    \[
    68 + 7a + b \quad \text{where} \quad 0 \leq 7a + b \leq \frac{3}{2}
    \]

11. The possible integer values of \( P(4) \) are:
    \[
    68, 69, 70, 71
    \]

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