Let $f : \mathbb{C} \to \mathbb{C} $ be defined by $ f(z) = z^2 + iz + 1 $. How many complex numbers $z $ are there such that $ \text{Im}(z) > 0 $ and both the real and the imaginary parts of $f(z)$ are integers with absolute value at most $ 10 $?

${ \textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D}} \ 431 \qquad \textbf{(E)} \ 441 $
1. Let \( f(z) = z^2 + iz + 1 \) where \( z = x + yi \) with \( x, y \in \mathbb{R} \) and \( y > 0 \). We need to find the number of complex numbers \( z \) such that both the real and imaginary parts of \( f(z) \) are integers with absolute value at most 10.

2. Express \( f(z) \) in terms of \( x \) and \( y \):
   \[
   f(z) = (x + yi)^2 + i(x + yi) + 1 = x^2 - y^2 + 2xyi + ix - y + 1
   \]
   \[
   f(z) = (x^2 - y^2 + x + 1) + i(2xy + y)
   \]
   Let \( f(z) = a + bi \) where \( a \) and \( b \) are integers. Then:
   \[
   x^2 - y^2 + x + 1 = a \quad \text{and} \quad 2xy + y = b
   \]

3. Solve for \( y \) in terms of \( x \) and \( b \):
   \[
   y(2x + 1) = b \implies y = \frac{b}{2x + 1}
   \]
   Since \( y > 0 \), \( \frac{b}{2x + 1} > 0 \). This implies \( b \) and \( 2x + 1 \) must have the same sign.

4. Substitute \( y = \frac{b}{2x + 1} \) into the real part equation:
   \[
   x^2 - \left(\frac{b}{2x + 1}\right)^2 + x + 1 = a
   \]
   \[
   x^2 - \frac{b^2}{(2x + 1)^2} + x + 1 = a
   \]

5. To find the number of valid pairs \((a, b)\), we need to ensure that both \( a \) and \( b \) are integers with \( |a| \leq 10 \) and \( |b| \leq 10 \). This gives us \( 21 \) possible values for each of \( a \) and \( b \) (from \(-10\) to \(10\)).

6. We need to count the number of pairs \((a, b)\) such that the imaginary part of \( f(z) \) is positive:
   \[
   \text{Im}\left(\sqrt{4a-5+4bi}\right) - 1 > 0
   \]
   Simplifying, we get:
   \[
   \sqrt{\frac{\sqrt{(4a-5)^2+(4b)^2}-(4a-5)}{2}} > 1
   \]
   \[
   \sqrt{(4a-5)^2 + (4b)^2} - (4a-5) > 2
   \]
   \[
   (4a-5)^2 + (4b)^2 > (4a-5)^2 + 4(4a-5) + 4
   \]
   \[
   4b^2 > 4a - 4 \implies b^2 > a - 1
   \]

7. Count the number of valid pairs \((a, b)\) for each range of \( a \):
   - For \( -10 \leq a \leq 0 \), there are \( 21 \) choices of \( b \) for each \( a \).
   - For \( a = 1 \), there are \( 20 \) choices of \( b \).
   - For \( 2 \leq a \leq 4 \), there are \( 18 \) choices of \( b \).
   - For \( 5 \leq a \leq 9 \), there are \( 16 \) choices of \( b \).
   - For \( a = 10 \), there are \( 14 \) choices of \( b \).

8. Calculate the total number of pairs:
   \[
   11 \times 21 + 1 \times 20 + 3 \times 18 + 5 \times 16 + 1 \times 14 = 231 + 20 + 54 + 80 + 14 = 399
   \]

Conclusion:
\(\boxed{399}\)