Let $$S = \{4, 6, 9\}$$ and $$T = \{9, 10, 11, \ldots, 1000\}$$. If $$A = \{a_1 + a_2 + \ldots + a_k : k \in \mathbb{N}, a_1, a_2, a_3, \ldots, a_k \in S\}$$, then the sum of all the elements in the set $$T - A$$ is equal to _______

We have $$S = \{4, 6, 9\}$$ and $$T = \{9, 10, 11, \ldots, 1000\}$$. The set $$A$$ consists of all numbers that can be represented as $$a_1 + a_2 + \cdots + a_k$$ where each $$a_i \in \{4, 6, 9\}$$ and $$k \in \mathbb{N}$$. We need to find the sum of all elements in $$T \setminus A$$, i.e., elements of $$T$$ that cannot be expressed as such sums.

We note that every element of $$A$$ is a non-negative integer combination of 4, 6, and 9 (with at least one term). The elements of $$A$$ starting from the smallest are: 4, 6, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, ... We need to find which numbers from 9 onwards are NOT in $$A$$.

We systematically check small values. A number $$n$$ is in $$A$$ if $$n = 4a + 6b + 9c$$ for non-negative integers $$a, b, c$$ with $$a + b + c \geq 1$$:

$$n = 4$$: $$4 = 4(1)$$. Yes. $$n = 6$$: $$6 = 6(1)$$. Yes. $$n = 8$$: $$8 = 4(2)$$. Yes. $$n = 9$$: $$9 = 9(1)$$. Yes. $$n = 10$$: $$10 = 4+6$$. Yes. $$n = 12$$: $$12 = 4(3)$$ or $$6(2)$$. Yes. $$n = 13$$: $$13 = 4+9$$. Yes. $$n = 14$$: $$14 = 4+4+6$$. Yes. $$n = 15$$: $$15 = 6+9$$. Yes. $$n = 16$$: $$16 = 4(4)$$. Yes. $$n = 17$$: $$17 = 4+4+9$$. Yes. $$n = 18$$: $$18 = 9(2)$$ or $$4(3)+6$$. Yes.

Now, $$n = 19$$: $$19 = 4+6+9$$. Yes. $$n = 20$$: $$20 = 4(5)$$. Yes. For $$n = 21$$: $$21 = 12+9 = 4(3)+9$$. Yes. $$n = 22$$: $$22 = 4+9+9 = 4+18$$. Yes. $$n = 23$$: $$23 = 4+4+6+9 = 14+9$$. Yes.

We observe that all integers $$\geq 24$$ can be written as sums: if $$n \equiv 0 \pmod{4}$$, use $$4$$'s; if $$n \equiv 1 \pmod{4}$$, use $$9 + 4$$'s (since $$n \geq 25$$ gives enough room, and $$n = 9$$ works, $$n = 13$$ works); if $$n \equiv 2 \pmod{4}$$, use $$6 + 4$$'s; if $$n \equiv 3 \pmod{4}$$, use $$9 + 6 + 4$$'s.

More precisely, for $$n \geq 24$$: $$n \equiv 0 \pmod{2}$$ can always be written as $$4a + 6b$$; $$n$$ odd and $$n \geq 9$$ can be written as $$9 + \text{even number} \geq 0$$. Since $$n - 9 \geq 15$$ for $$n \geq 24$$, and $$n - 9$$ is even, and any even number $$\geq 4$$ can be written as $$4a + 6b$$ (and even numbers $$\geq 4$$: 4, 6, 8, 10, ... all work). Actually even number 0 also works (take nothing extra). So for odd $$n \geq 9$$: $$n - 9$$ is even and $$\geq 0$$, and we need $$n - 9 \neq 2$$ (since 2 cannot be made from 4s and 6s). $$n - 9 = 2 \Rightarrow n = 11$$. So $$n = 11$$ is the concern.

Let us check $$n = 11$$: Can we write $$11 = 4a + 6b + 9c$$? With $$c = 1$$: $$2 = 4a + 6b$$, no non-negative solution. With $$c = 0$$: $$11 = 4a + 6b$$, impossible (parity: $$4a + 6b$$ is even, 11 is odd). So $$n = 11 \notin A$$.

We also need to check numbers less than 24 more carefully. From the set $$T = \{9, 10, \ldots, 1000\}$$, which values from 9 to 23 are NOT in $$A$$?

$$n = 9$$: Yes (checked). $$n = 10$$: Yes. $$n = 11$$: No (shown above). $$n = 12$$: Yes. $$n = 13$$: Yes. $$n = 14$$: Yes. $$n = 15$$: Yes. $$n = 16$$: Yes. $$n = 17$$: Yes. $$n = 18$$: Yes. $$n = 19$$: Yes. $$n = 20$$: Yes. $$n = 21$$: Yes. $$n = 22$$: Yes. $$n = 23$$: Yes.

For $$n \geq 24$$, every number is representable as argued. So the only element of $$T$$ not in $$A$$ is $$n = 11$$.

Therefore, the sum of all elements in $$T \setminus A$$ is $$\boxed{11}$$.

Hence, the correct answer is 11.