Given the relation $R$ on $A = \{1, 2, 3, \ldots, 100\}$ defined by $(x, y) \in R$ if and only if $2x = 3y$.

We need to find the pairs $(x, y)$ such that $2x = 3y$ and $x, y \in A$.

Since $2x = 3y$, we have $x = \frac{3}{2}y$. Since $x$ is an integer, $y$ must be an even number.

Let $y = 2k$, where $k$ is an integer. Then $x = \frac{3}{2}(2k) = 3k$.

Since $x, y \in A$, we have $1 \le x \le 100$ and $1 \le y \le 100$.

So, $1 \le 3k \le 100$ and $1 \le 2k \le 100$.

This gives $1 \le k \le \frac{100}{3} = 33.33$ and $1 \le k \le \frac{100}{2} = 50$.

Therefore, $1 \le k \le 33$.

The pairs in $R$ are of the form $(3k, 2k)$ for $k = 1, 2, \ldots, 33$.

So, $R = \{(3, 2), (6, 4), (9, 6), \ldots, (99, 66)\}$. The number of elements in $R$ is 33.

Now, $R_1$ is a symmetric relation such that $R \subset R_1$. For $R_1$ to be symmetric, if $(x, y) \in R_1$, then $(y, x) \in R_1$.

Since $R \subset R_1$, all elements of $R$ must be in $R_1$. Also, for each $(x, y) \in R$, $(y, x)$ must also be in $R_1$ to make it symmetric.

So, if $(3k, 2k) \in R$, then $(2k, 3k)$ must be in $R_1$.

We need to check if $(2k, 3k) \in A \times A$, i.e., $1 \le 2k \le 100$ and $1 \le 3k \le 100$.

This gives $1 \le k \le 50$ and $1 \le k \le 33.33$. So, $1 \le k \le 33$.

The elements in $R_1$ will be the elements of $R$ and their symmetric counterparts. So, $R_1$ will contain $(3k, 2k)$ and $(2k, 3k)$ for $k = 1, 2, \ldots, 33$.

The number of elements in $R_1$ will be $33 + 33 = 66$.

However, we also need to consider the elements of the form $(x, x)$ in $R_1$ to ensure it is symmetric. Since $2x = 3y$, for $(x, x)$ to be in $R$, we need $2x = 3x$, which implies $x = 0$. But $0 \notin A$. So, there are no elements of the form $(x, x)$ in $R$.

To make $R_1$ symmetric, we need to add $(2k, 3k)$ for each $(3k, 2k)$ in $R$.

So, $R_1 = \{(3k, 2k) : k = 1, 2, \ldots, 33\} \cup \{(2k, 3k) : k = 1, 2, \ldots, 33\}$.

The number of elements in $R_1$ is $33 + 33 = 66$.

However, we can also have elements of the form $(x, x)$ in $R_1$. Since $R \subset R_1$, we need to add the minimum number of elements to $R_1$ to make it symmetric.

The number of elements in $R_1$ is $n$. We want to minimize $n$.

The elements in $R$ are $(3k, 2k)$ for $k = 1, 2, \ldots, 33$. To make $R_1$ symmetric, we need to add $(2k, 3k)$ for $k = 1, 2, \ldots, 33$.

So, $R_1$ contains $(3k, 2k)$ and $(2k, 3k)$ for $k = 1, 2, \ldots, 33$. The number of elements is $33 + 33 = 66$.

If we add elements of the form $(x, x)$, we need to add $(1, 1), (2, 2), \ldots, (100, 100)$. But these are not required to make $R_1$ symmetric since $R$ does not contain any such elements.

Therefore, the minimum value of $n$ is 66.