Let $$A = \{1, a_1, a_2, \ldots a_{18}, 77\}$$ be a set of integers with $$1 < a_1 < a_2 < \ldots < a_{18} < 77$$. Let the set $$A + A = \{x + y : x, y \in A\}$$ contain exactly $$39$$ elements. Then, the value of $$a_1 + a_2 + \ldots + a_{18}$$ is equal to ______

$$A = \{1, a_1, a_2, \ldots, a_{18}, 77\}$$ has 20 elements, and $$A + A$$ has exactly 39 elements.

For any set of $$n$$ elements, $$|A + A| \geq 2n - 1$$. Here $$n = 20$$, so $$|A + A| \geq 39$$. Since $$|A + A| = 39$$ achieves this minimum, $$A$$ must be an arithmetic progression.

With 20 terms, first term 1, last term 77:

$$77 = 1 + 19d \implies d = \frac{76}{19} = 4$$

So $$A = \{1, 5, 9, 13, \ldots, 77\}$$, an AP with $$a_i = 1 + 4i$$ for $$i = 1, 2, \ldots, 18$$.

$$a_1 + a_2 + \ldots + a_{18} = \sum_{i=1}^{18} (1 + 4i) = 18 + 4 \cdot \frac{18 \times 19}{2} = 18 + 684 = 702$$

Hence the answer is $$\boxed{702}$$.