Let $$A = \{-3, -2, -1, 0, 1, 2, 3\}$$. Let R be a relation on A defined by $$xRy$$ if and only if $$0 \leq x^2 + 2y \leq 4$$. Let $$l$$ be the number of elements in R and $$m$$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $$l + m$$ is equal to

We have $$A = \{-3, -2, -1, 0, 1, 2, 3\}$$ and the relation $$R$$ on $$A$$ defined by $$xRy$$ if and only if $$0 \leq x^2 + 2y \leq 4$$.

For a given $$x$$, the condition $$0 \leq x^2 + 2y \leq 4$$ gives us:
$$-x^2 \leq 2y \leq 4 - x^2$$
$$\frac{-x^2}{2} \leq y \leq \frac{4 - x^2}{2}$$

We check each value of $$x$$ in $$A$$:

Case 1: $$x = -3$$ (so $$x^2 = 9$$)
We need $$0 \leq 9 + 2y \leq 4$$, which gives $$-4.5 \leq y \leq -2.5$$.
From $$A$$, the valid values are $$y \in \{-3\}$$.
Pairs: $$(-3, -3)$$. Count = 1.

Case 2: $$x = -2$$ (so $$x^2 = 4$$)
We need $$0 \leq 4 + 2y \leq 4$$, which gives $$-2 \leq y \leq 0$$.
From $$A$$, the valid values are $$y \in \{-2, -1, 0\}$$.
Pairs: $$(-2, -2), (-2, -1), (-2, 0)$$. Count = 3.

Case 3: $$x = -1$$ (so $$x^2 = 1$$)
We need $$0 \leq 1 + 2y \leq 4$$, which gives $$-0.5 \leq y \leq 1.5$$.
From $$A$$, the valid values are $$y \in \{0, 1\}$$.
Pairs: $$(-1, 0), (-1, 1)$$. Count = 2.

Case 4: $$x = 0$$ (so $$x^2 = 0$$)
We need $$0 \leq 2y \leq 4$$, which gives $$0 \leq y \leq 2$$.
From $$A$$, the valid values are $$y \in \{0, 1, 2\}$$.
Pairs: $$(0, 0), (0, 1), (0, 2)$$. Count = 3.

Case 5: $$x = 1$$ (so $$x^2 = 1$$)
We need $$0 \leq 1 + 2y \leq 4$$, which gives $$-0.5 \leq y \leq 1.5$$.
From $$A$$, the valid values are $$y \in \{0, 1\}$$.
Pairs: $$(1, 0), (1, 1)$$. Count = 2.

Case 6: $$x = 2$$ (so $$x^2 = 4$$)
We need $$0 \leq 4 + 2y \leq 4$$, which gives $$-2 \leq y \leq 0$$.
From $$A$$, the valid values are $$y \in \{-2, -1, 0\}$$.
Pairs: $$(2, -2), (2, -1), (2, 0)$$. Count = 3.

Case 7: $$x = 3$$ (so $$x^2 = 9$$)
We need $$0 \leq 9 + 2y \leq 4$$, which gives $$-4.5 \leq y \leq -2.5$$.
From $$A$$, the valid values are $$y \in \{-3\}$$.
Pairs: $$(3, -3)$$. Count = 1.

So the total number of elements in $$R$$ is:
$$l = 1 + 3 + 2 + 3 + 2 + 3 + 1 = 15$$

Now we find $$m$$, the minimum number of elements to be added to make $$R$$ reflexive. For reflexivity, we need $$(x, x) \in R$$ for every $$x \in A$$. Let us check which diagonal pairs are already in $$R$$:

$$(-3, -3)$$: $$9 + 2(-3) = 3$$. Since $$0 \leq 3 \leq 4$$, this is in $$R$$. ✓
$$(-2, -2)$$: $$4 + 2(-2) = 0$$. Since $$0 \leq 0 \leq 4$$, this is in $$R$$. ✓
$$(-1, -1)$$: $$1 + 2(-1) = -1$$. Since $$-1 \lt 0$$, this is NOT in $$R$$. ✗
$$(0, 0)$$: $$0 + 2(0) = 0$$. Since $$0 \leq 0 \leq 4$$, this is in $$R$$. ✓
$$(1, 1)$$: $$1 + 2(1) = 3$$. Since $$0 \leq 3 \leq 4$$, this is in $$R$$. ✓
$$(2, 2)$$: $$4 + 2(2) = 8$$. Since $$8 \gt 4$$, this is NOT in $$R$$. ✗
$$(3, 3)$$: $$9 + 2(3) = 15$$. Since $$15 \gt 4$$, this is NOT in $$R$$. ✗

We need to add 3 pairs: $$(-1, -1), (2, 2), (3, 3)$$. So $$m = 3$$.

Therefore, $$l + m = 15 + 3 = 18$$.

Hence, the correct answer is Option D.