1. We need all ordered pairs $(x,y) \in A\times A$ such that $2x=3y,$ where $A=\{1,2,3,\dots,100\}.$

2. Solve the relation condition. Since $2x=3y$, we must have $x$ divisible by $3$ and $y$ divisible by $2$. Let $x=3k,\quad y=2k$ for some positive integer $k$.

Because $x,y\in A$, $3k\le 100,\qquad 2k\le 100.$ The stricter condition is $k\le \left\lfloor \frac{100}{3}\right\rfloor=33.$ So the pairs in $R$ are $(3k,2k),\quad k=1,2,3,\dots,33.$ Hence, $|R|=33.$

3. Now $R_1$ must be a symmetric relation containing $R$. A relation is symmetric if whenever $(a,b)\in R_1$, then $(b,a)\in R_1$ also.

So for every pair $(3k,2k)\in R$, we must also include $(2k,3k).$

4. Check whether any pair in $R$ is already symmetric by itself, i.e. of the form $(a,a)$. That would require $3k=2k \implies k=0,$ which is not allowed since $k\ge 1$. So no pair in $R$ is of the form $(a,a)$.

Also, $(3k,2k)\ne (2k,3k)$ for all $k\ge 1$. Thus each pair in $R$ contributes one new distinct reverse pair.

5. Therefore the smallest symmetric relation containing $R$ is $R_1=R\cup R^{-1},$ and since $R\cap R^{-1}=\varnothing$, $|R_1|=|R|+|R^{-1}|=33+33=66.$

6. Hence the minimum value of $n$ is $\boxed{66}.$