1. Let the speed of the body just before hitting the ground be $v_1$ and just after rebound be $v_2$.

2. Given: $\frac{v_1}{v_2}=4 \implies v_1=4v_2$

3. Kinetic energy just before collision: $K_1=\frac{1}{2}mv_1^2$

4. Kinetic energy just after collision: $K_2=\frac{1}{2}mv_2^2$

Using $v_1=4v_2$, $K_1=\frac{1}{2}m(4v_2)^2=16\left(\frac{1}{2}mv_2^2\right)=16K_2$

So, $\frac{K_2}{K_1}=\frac{1}{16}$

5. Loss in kinetic energy: $\Delta K=K_1-K_2$

Percentage loss: $\frac{K_1-K_2}{K_1}\times 100=\left(1-\frac{1}{16}\right)\times 100$ $=\frac{15}{16}\times 100=93.75\%$

6. Given that percentage loss is $\frac{x}{4}$, $\frac{x}{4}=93.75$

Thus, $x=93.75\times 4=375$

Therefore, the required integer is: $\boxed{375}$