There are relatively prime positive integers $m$ and $n$ so that the parabola with equation $y = 4x^2$ is tangent to the parabola with equation $x = y^2 + \frac{m}{n}$ . Find $m + n$.
1. Let \(\frac{m}{n} = e\). We substitute \(y = 4x^2\) into \(x = y^2 + e\) to obtain:
   \[
   x = (4x^2)^2 + e \implies x = 16x^4 + e
   \]
2. We want the quartic polynomial \(16x^4 - x + e\) to have a repeated real root, which implies that the polynomial must have a discriminant of zero. The discriminant \(\Delta\) of a quartic polynomial \(ax^4 + bx^3 + cx^2 + dx + e\) is given by:
   \[
   \Delta = 256a^3e^3 - 192a^2bde^2 - 128a^2c^2e^2 + 144a^2cd^2e - 27a^2d^4 + 144ab^2ce^2 - 6ab^2d^2e - 80abc^2de + 18abcd^3 + 16ac^4e - 4ac^3d^2 - 27b^4e^2 + 18b^3cde - 4b^3d^3 - 4b^2c^3e + b^2c^2d^2
   \]
3. For our polynomial \(16x^4 - x + e\), we have \(a = 16\), \(b = 0\), \(c = 0\), \(d = -1\), and \(e = e\). Substituting these values into the discriminant formula, we focus on the terms that only involve \(a\), \(d\), and \(e\):
   \[
   \Delta = 256a^3e^3 - 27a^2d^4
   \]
4. Plugging in \(a = 16\) and \(d = -1\), we get:
   \[
   \Delta = 256(16^3)e^3 - 27(16^2)(-1)^4 = 256 \cdot 4096 \cdot e^3 - 27 \cdot 256 = 1048576e^3 - 6912
   \]
5. Setting the discriminant to zero for the polynomial to have a repeated root:
   \[
   1048576e^3 - 6912 = 0 \implies 1048576e^3 = 6912 \implies e^3 = \frac{6912}{1048576} = \frac{27}{4096} \implies e = \sqrt[3]{\frac{27}{4096}} = \frac{3}{16}
   \]
6. Thus, \(e = \frac{3}{16}\), so \(m = 3\) and \(n = 16\). Therefore, \(m + n = 3 + 16 = 19\).

The final answer is \(\boxed{19}\)