Let $ABC$ be a triangle and $\Gamma$ the $A$- exscribed circle whose center is $J$ . Let $D$ and $E$ be the touchpoints of $\Gamma$  with the lines $AB$ and $AC$, respectively. Let $S$ be the area of the quadrilateral $ADJE$, Find the maximum value that $\frac{S}{AJ^2}$ has and when equality holds.
1. **Define the given elements and their relationships:**
   - Let $ABC$ be a triangle.
   - $\Gamma$ is the $A$-excircle with center $J$.
   - $D$ and $E$ are the touchpoints of $\Gamma$ with $AB$ and $AC$, respectively.
   - $S$ is the area of the quadrilateral $ADJE$.

2. **Determine the radius of the $A$-excircle:**
   - The radius of the $A$-excircle, denoted as $r_A$, is given by:
     \[
     r_A = JD = s \cdot \tan \frac{\alpha}{2}
     \]
     where $s$ is the semiperimeter of $\triangle ABC$ and $\alpha$ is the angle at vertex $A$.

3. **Calculate the length of $AD$ and $AJ$:**
   - $AD = s$ (since $D$ is the touchpoint of the $A$-excircle with $AB$).
   - $AJ = \frac{s}{\cos \frac{\alpha}{2}}$ (using the relationship between the exradius and the semiperimeter).

4. **Express the area $S$ of quadrilateral $ADJE$:**
   - The area $S$ of quadrilateral $ADJE$ can be calculated as:
     \[
     S = AD \cdot JD = s \cdot s \cdot \tan \frac{\alpha}{2} = s^2 \cdot \tan \frac{\alpha}{2}
     \]

5. **Calculate $\frac{S}{AJ^2}$:**
   - Substitute the expressions for $S$ and $AJ$:
     \[
     \frac{S}{AJ^2} = \frac{s^2 \cdot \tan \frac{\alpha}{2}}{\left(\frac{s}{\cos \frac{\alpha}{2}}\right)^2}
     \]
   - Simplify the expression:
     \[
     \frac{S}{AJ^2} = \frac{s^2 \cdot \tan \frac{\alpha}{2}}{\frac{s^2}{\cos^2 \frac{\alpha}{2}}} = \tan \frac{\alpha}{2} \cdot \cos^2 \frac{\alpha}{2}
     \]
   - Use the trigonometric identity $\tan x = \frac{\sin x}{\cos x}$:
     \[
     \frac{S}{AJ^2} = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \cdot \cos^2 \frac{\alpha}{2} = \sin \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2}
     \]
   - Simplify further using the double-angle identity $\sin 2x = 2 \sin x \cos x$:
     \[
     \frac{S}{AJ^2} = \frac{1}{2} \sin \alpha
     \]

6. **Determine the maximum value of $\frac{S}{AJ^2}$:**
   - The maximum value of $\sin \alpha$ is $1$, which occurs when $\alpha = 90^\circ$.
   - Therefore, the maximum value of $\frac{S}{AJ^2}$ is:
     \[
     \frac{1}{2} \sin 90^\circ = \frac{1}{2} \cdot 1 = \frac{1}{2}
     \]

Conclusion:
\[
\boxed{\frac{1}{2}}
\]