$$\begin{gathered} \mathrm{If}C_{r+1}^{36}=6\cdot \frac{C_r^{35}}{k^2-3}, \\ \mathrm{then}\mathrm{number}\mathrm{of}\mathrm{ordered}\mathrm{pairs}(r,k), \\ \mathrm{where}r,k\in \mathbb{Z}, \\ \mathrm{is}\_\_\_\_ \end{gathered}$$

The correct answer is 4.

Let us write down the given equation:
$C_{r+1}^{36}=6\cdot \frac{C_r^{35}}{k^2-3}$

For the combinations $C_{r+1}^{36}$ and $C_r^{35}$ to be defined and non-zero, we must have:
$0\le r\le 35$ and $0\le r+1\le 36$
Since $r\in \mathbb{Z}$, this gives the domain for $r$ as:
$r\in \{0,1,2,\cdots ,35\}$

Using the combination identity $C_y^x=\frac{x}{y}\cdot C_{y-1}^{x-1}$, we can write:
$C_{r+1}^{36}=\frac{36}{r+1}\cdot C_r^{35}$

Substitute this back into the original equation:
$\frac{36}{r+1}\cdot C_r^{35}=6\cdot \frac{C_r^{35}}{k^2-3}$

Since $C_{\mathrm{r<!--<mn-->35}}\ne 0$, we can divide both sides by $C_r^{35}$:
$\frac{36}{r+1}=\frac{6}{k^2-3}$

Dividing both sides by 6, we get:
$\frac{6}{r+1}=\frac{1}{k^2-3}$
Rearranging for $r+1$:
$r+1=6(k^2-3)$

Since we require $0\le r\le 35$, we have:
$1\le r+1\le 36$
Substituting $r+1=6(k^2-3)$ into the inequality:
$1\le 6(k^2-3)\le 36$
Dividing by 6:
$\frac{1}{6}\le k^2-3\le 6$
Adding 3 to all parts of the inequality:
$3\frac{1}{6}\le k^2\le 9$

Since $k\in \mathbb{Z}$, its square $k^2$ must be a perfect square integer. The only perfect square integers between $3.17$ and $9$ inclusive are $4$ and $9$.

Let's analyze these cases:
Case 1: $k^2=4$
This gives $k=2$ or $k=-2$.
Substituting $k^2=4$ into the equation for $r$:
$r+1=6(4-3)=6\Rightarrow r=5$
Since $0\le 5\le 35$, this yields two valid ordered pairs: $(5,2)$ and $(5,-2)$.

Case 2: $k^2=9$
This gives $k=3$ or $k=-3$.
Substituting $k^2=9$ into the equation for $r$:
$r+1=6(9-3)=36\Rightarrow r=35$
Since $0\le 35\le 35$, this yields another two valid ordered pairs: $(35,3)$ and $(35,-3)$.

Thus, the complete set of ordered pairs $(r,k)$ is:
$\left\{\right(5,2),(5,-2),(35,3),(35,-3\left)\right\}$

The total number of ordered pairs is 4.