1. Set up the GP

Let the first term be $a$ and common ratio be $r=\frac{1}{m}, \qquad m\in \mathbb{N}.$

Given that the 4th term is $500$: $ar^3=500.$ So, $a\left(\frac{1}{m}\right)^3=500 \implies a=500m^3.$

Hence the GP is $500m^3,\; 500m^2,\; 500m,\; 500,\; \frac{500}{m},\; \frac{500}{m^2},\; \frac{500}{m^3},\dots$

2. Use the condition $S_6>S_5+1$

Since $S_6-S_5=T_6,$ we get $T_6>1.$ Now, $T_6=ar^5=500m^3\left(\frac{1}{m}\right)^5=\frac{500}{m^2}.$ Thus, $\frac{500}{m^2}>1 \implies m^2<500.$ So, $m<\sqrt{500}\approx 22.36.$ Since $m\in\mathbb N$, $m\le 22.$

3. Use the condition $S_7<S_6+\frac12$

Similarly, $S_7-S_6=T_7,$ so $T_7<\frac12.$ Now, $T_7=ar^6=500m^3\left(\frac{1}{m}\right)^6=\frac{500}{m^3}.$ Thus, $\frac{500}{m^3}<\frac12.$ Multiply both sides by $2m^3$: $1000<m^3.$ So, $m>10.$ Since $m$ is a natural number, $m\ge 11.$

4. Combine both conditions

We need $11\le m\le 22.$ So the possible values are $m=11,12,13,\dots,22.$

Number of integers in this range: $22-11+1=12.$

5. Final answer

The number of possible values of $m$ is $\boxed{12}.$