Let $A, B$ and $C$ be three positive integers such that the sum of $A$ and the mean of $B$ and $C$ is $5.$ In addition, the sum of $B$ and the mean of $A$ and $C$ is $7.$ Then the sum of $A$ and $B$ is

The mean of two numbers is their average. So the mean of B and C is (B + C)/2.

We translate the given conditions into mathematical equations.

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The sum of A and the mean of B and C is 5:

$A + \dfrac{B + C}{2} = 5$

The sum of B and the mean of A and C is 7:

$B + \dfrac{A + C}{2} = 7$

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To make these equations easier to work with, we multiply each equation by 2:

From the first condition:

$2A + B + C = 10$ ...(1)

From the second condition:

$2B + A + C = 14$ ...(2)

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Now we have a clean system of equations. Subtracting equation (1) from equation (2):

$(2B + A + C) - (2A + B + C) = 14 - 10$

$2B + A + C - 2A - B - C = 4$

$B - A = 4$ ...(3)

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From equation (3), we know that $B = A + 4$.

Substituting this back into equation (1):

$2A + (A + 4) + C = 10$

$3A + C = 6$ ...(4)

Substituting into equation (2):

$2(A + 4) + A + C = 14$

$2A + 8 + A + C = 14$

$3A + C = 6$ ...(5)

Both equations give us the same result: $3A + C = 6$.

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Since A, B, and C are positive integers, and $C = 6 - 3A$, we need:

$C > 0$, which means $6 - 3A > 0$, so $A < 2$

Since A is a positive integer, $A = 1$

Therefore:

$A = 1$

$B = A + 4 = 5$

$C = 6 - 3(1) = 3$

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Therefore, $A + B = 1 + 5 = 6$.

The beauty of this problem is that once we found $B - A = 4$, we could determine that there's only one possible solution where all three numbers are positive integers. This constraint significantly limits our possibilities and leads us directly to the answer.