A function $ f$ is defined by $ f(z) \equal{} (4 \plus{} i) z^2 \plus{} \alpha z \plus{} \gamma$ for all complex numbers $ z$, where $ \alpha$ and $ \gamma$ are complex numbers and $ i^2 \equal{} \minus{} 1$. Suppose that $ f(1)$ and $ f(i)$ are both real. What is the smallest possible value of $ | \alpha | \plus{} |\gamma |$?

$ \textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad$
1. **Define the function and given conditions:**
   The function \( f(z) = (4 + i)z^2 + \alpha z + \gamma \) is given, where \( \alpha \) and \( \gamma \) are complex numbers. We know that \( f(1) \) and \( f(i) \) are both real numbers.

2. **Express \( \alpha \) and \( \gamma \) in terms of real and imaginary parts:**
   Let \( \alpha = a + bi \) and \( \gamma = c + di \), where \( a, b, c, \) and \( d \) are real numbers.

3. **Calculate \( f(1) \):**
   \[
   f(1) = (4 + i) \cdot 1^2 + \alpha \cdot 1 + \gamma = 4 + i + a + bi + c + di = (4 + a + c) + (1 + b + d)i
   \]
   Since \( f(1) \) is real, the imaginary part must be zero:
   \[
   1 + b + d = 0 \implies b + d = -1
   \]

4. **Calculate \( f(i) \):**
   \[
   f(i) = (4 + i) \cdot i^2 + \alpha \cdot i + \gamma = (4 + i) \cdot (-1) + (a + bi) \cdot i + c + di
   \]
   Simplify the expression:
   \[
   f(i) = -4 - i + ai - b + c + di = (-4 - b + c) + (a + d - 1)i
   \]
   Since \( f(i) \) is real, the imaginary part must be zero:
   \[
   a + d - 1 = 0 \implies a + d = 1
   \]

5. **Solve the system of equations:**
   We have two equations:
   \[
   b + d = -1
   \]
   \[
   a + d = 1
   \]
   Subtract the first equation from the second:
   \[
   (a + d) - (b + d) = 1 - (-1) \implies a - b = 2
   \]

6. **Minimize \( |\alpha| + |\gamma| \):**
   We need to minimize \( |\alpha| + |\gamma| = \sqrt{a^2 + b^2} + \sqrt{c^2 + d^2} \).

7. **Choose \( c = 0 \) to simplify:**
   Since there are no restrictions on \( c \), we can set \( c = 0 \). Thus, \( \gamma = di \).

8. **Solve for \( d \):**
   From \( b + d = -1 \) and \( a + d = 1 \), we have:
   \[
   d = 1 - a
   \]
   \[
   b = -1 - d = -1 - (1 - a) = -2 + a
   \]

9. **Substitute \( a \) and \( b \) into the equation \( a - b = 2 \):**
   \[
   a - (-2 + a) = 2 \implies a + 2 - a = 2 \implies 2 = 2
   \]
   This confirms our values are consistent.

10. **Calculate \( |\alpha| \) and \( |\gamma| \):**
    \[
    |\alpha| = \sqrt{a^2 + b^2} = \sqrt{a^2 + (-2 + a)^2} = \sqrt{a^2 + 4 - 4a + a^2} = \sqrt{2a^2 - 4a + 4}
    \]
    \[
    |\gamma| = |d| = |1 - a|
    \]

11. **Minimize the expression:**
    To minimize \( \sqrt{2a^2 - 4a + 4} + |1 - a| \), we test \( a = 1 \) and \( b = -1 \):
    \[
    |\alpha| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}
    \]
    \[
    |\gamma| = |1 - 1| = 0
    \]
    Thus, \( |\alpha| + |\gamma| = \sqrt{2} + 0 = \sqrt{2} \).

The final answer is \( \boxed{\sqrt{2}} \).