1. Let $E=\tan 9^\circ-\tan 27^\circ-\tan 63^\circ+\tan 81^\circ.$

2. Use the complementary-angle identity: $\tan(90^\circ-\theta)=\cot\theta=\frac{1}{\tan\theta}.$ So, $\tan 81^\circ=\cot 9^\circ=\frac{1}{\tan 9^\circ},$ $\tan 63^\circ=\cot 27^\circ=\frac{1}{\tan 27^\circ}.$

Thus, $E=\tan 9^\circ-\tan 27^\circ-\frac{1}{\tan 27^\circ}+\frac{1}{\tan 9^\circ}.$

3. Group the terms: $E=\left(\tan 9^\circ+\frac{1}{\tan 9^\circ}\right)-\left(\tan 27^\circ+\frac{1}{\tan 27^\circ}\right).$

Using $x+\frac{1}{x}=\frac{x^2+1}{x}=\frac{\sec^2\theta}{\tan\theta}=\frac{1}{\sin\theta\cos\theta}=\frac{2}{\sin 2\theta},$ we get $\tan\theta+\cot\theta=\frac{2}{\sin 2\theta}.$

Hence, $E=\frac{2}{\sin 18^\circ}-\frac{2}{\sin 54^\circ}.$

4. Use exact values: $\sin 18^\circ=\frac{\sqrt5-1}{4},$ $\sin 54^\circ=\cos 36^\circ=\frac{\sqrt5+1}{4}.$

So, $E=\frac{2}{(\sqrt5-1)/4}-\frac{2}{(\sqrt5+1)/4}$ $=\frac{8}{\sqrt5-1}-\frac{8}{\sqrt5+1}.$

5. Simplify: $E=8\left(\frac{1}{\sqrt5-1}-\frac{1}{\sqrt5+1}\right)$ $=8\left(\frac{(\sqrt5+1)-(\sqrt5-1)}{(\sqrt5-1)(\sqrt5+1)}\right)$ $=8\left(\frac{2}{5-1}\right)$ $=8\cdot \frac12=4.$

6. Therefore, the required integer value is $\boxed{4}.$