1. Given relation

We have $A=\{0,3,4,6,7,8,9,10\}$ and $R=\{(x,y)\in A\times A: x-y \text{ is a positive odd integer or } x-y=2\}.$

We must add the minimum number of ordered pairs so that $R$ becomes symmetric.

A relation is symmetric if $(x,y)\in R \implies (y,x)\in R.$

So we first count all pairs already in $R$, and then count how many reverse pairs are missing.

2. List all elements in increasing order

$0,3,4,6,7,8,9,10$

Now for each $x\in A$, include all $y\in A$ such that

• $x-y$ is a positive odd integer, or
• $x-y=2$.

That means $x>y$, and allowed differences are $1,2,3,5,7,9,\dots$

3. Find all pairs in $R$

We check each first coordinate $x$.

For $x=0$

No smaller element in $A$, so no pair.

For $x=3$

Possible differences with elements of $A$:

• $3-0=3$ (positive odd) $\Rightarrow (3,0)$

So: $(3,0)$

For $x=4$

• $4-3=1$ (positive odd) $\Rightarrow (4,3)$
• $4-0=4$ not allowed

So: $(4,3)$

For $x=6$

• $6-4=2$ allowed $\Rightarrow (6,4)$
• $6-3=3$ odd $\Rightarrow (6,3)$
• $6-0=6$ not allowed

So: $(6,4),(6,3)$

For $x=7$

• $7-6=1$ odd $\Rightarrow (7,6)$
• $7-4=3$ odd $\Rightarrow (7,4)$
• $7-0=7$ odd $\Rightarrow (7,0)$
• $7-3=4$ not allowed

So: $(7,6),(7,4),(7,0)$

For $x=8$

• $8-6=2$ allowed $\Rightarrow (8,6)$
• $8-7=1$ odd $\Rightarrow (8,7)$
• $8-3=5$ odd $\Rightarrow (8,3)$
• $8-0=8$ not allowed
• $8-4=4$ not allowed

So: $(8,6),(8,7),(8,3)$

For $x=9$

• $9-8=1$ odd $\Rightarrow (9,8)$
• $9-7=2$ allowed $\Rightarrow (9,7)$
• $9-6=3$ odd $\Rightarrow (9,6)$
• $9-4=5$ odd $\Rightarrow (9,4)$
• $9-0=9$ odd $\Rightarrow (9,0)$
• $9-3=6$ not allowed

So: $(9,8),(9,7),(9,6),(9,4),(9,0)$

For $x=10$

• $10-9=1$ odd $\Rightarrow (10,9)$
• $10-8=2$ allowed $\Rightarrow (10,8)$
• $10-7=3$ odd $\Rightarrow (10,7)$
• $10-6=4$ not allowed
• $10-4=6$ not allowed
• $10-3=7$ odd $\Rightarrow (10,3)$
• $10-0=10$ not allowed

So: $(10,9),(10,8),(10,7),(10,3)$

4. Count elements of $R$

Total pairs: $1+1+2+3+3+5+4=19$

So $|R|=19.$

5. Check whether any reverse pair is already present

Every pair in $R$ satisfies $x>y$ because $x-y$ is positive odd or $2$. So no pair of the form $(y,x)$ can already be in $R$, since that would require $y>x$ and simultaneously satisfy the same condition.

Also, no diagonal pair $(x,x)$ belongs to $R$ because $x-x=0$, which is neither positive odd nor $2$.

Hence for every $(x,y)\in R$, the reverse pair $(y,x)$ is missing.

To make the relation symmetric, we must add all these reverse pairs. Therefore, the minimum number of pairs to be added is exactly $|R|=19.$

6. Final answer

$\boxed{19}$

The derived answer matches the stored correct answer.