$8, 9, 10, 13, 17, 19, 27$

Let $a$ be the sum of two numbers the player $A$ wrote, and $b$ be the sum of the two numbers $B$ wrote. If we let $t, s$ be the two numbers $A$ wrote, then $t+s = a$ and $ts = 2b$, so $t, s$ must be the solutions of the quadratic equation $x^2 - a x + 2b = 0$. Similarly, the two numbers $B$ wrote down are the solutions of the equation $x^2 - b x + 2a = 0$. From this we obtain the fact that the discriminants $a^2 - 8b$ and $b^2 - 8a$ of these equations must be perfect squares. Conversely, suppose that $a^2 - 8b$ is a perfect square, equal to $k^2$ for positive integer $k$, say. Then, we see that from $(a+k)(a-k) = 8b$ that $a > k$ and that $a$ and $k$ have the same even-odd parity. Therefore, the two roots $\frac{a+k}{2}$, $\frac{a-k}{2}$ of the equation $x^2 - a x + 2b = 0$ are positive integers. Similarly, if $b^2 - 8a$ is a perfect square, we see that the two roots of the equation $x^2 - b x + 2a = 0$ are positive integers. Thus, we see that the conditions for the problem are equivalent to the following:
$a$ and $b$ are positive integers and both $a^2-8b$ and $b^2-8a$ are perfect squares, and $a \ge b$.
In the sequel, we determine the pairs $(a, b)$ satisfying these conditions.
Since $a \ge b$, we see that if $a > 9$, then
$$a^2 > a^2 - 8b \geq a^2 - 8a > (a-6)^2$$
must hold. As $a^2 - 8b$ and $a$ have the same even-odd parity, we must have either $a^2 - 8b = (a-2)^2$ or $a^2 - 8b = (a-4)^2$. From the former we get $b = \frac{a-1}{2}$ and from the latter $b = a - 2$.
Now, suppose $b = \frac{a-1}{2}$ is satisfied. Then, we have $b^2 - 8a = b^2 - 8(2b + 1) = b^2 - 16b - 8 < (b-8)^2$. Since $b^2 - 8a$ and $b$ have the same even-odd parity, we see that we can write
$$b^2 - 16b - 8 = (b - 2\ell)^2,$$
where $\ell$ is an integer greater than 4. Then, we get
$$b = \frac{\ell^2+2}{2} = (\ell + 4) + \frac{18}{\ell-4}$$
and we conclude that $\ell = 5, 6, 7, 10, 13, 22$ must hold and these values of $\ell$ yield $b = 17, 19, 27$ with corresponding values for $a = 35, 39, 55$ respectively.
Next suppose $b = a-2$ is satisfied. Then, we have $b^2 - 8a - 16 < (b-4)^2$. Since $b^2 - 8a$ and $b$ have the same even-odd parity, we can write
$$b^2 - 8b - 16 = (b - 2j)^2,$$
where $j$ is an integer greater than 2. Then, we get
$$b = \frac{j^2+4}{j-2} = (j + 2) + \frac{8}{j-2},$$
from which we get $j = 3, 4, 6, 10$, which yield $b = 10, 13$ and $a = 12, 15$ respectively.
Finally, let us consider the case where $a \le 9$. If $b < 8$, then we get $b^2 - 8a \le b^2 - 8b < 0$, which does not satisfy our requirement. So, we must have $b = 8$ or $9$. It is easy to check that $(a,b) = (9,8)$ does not satisfy the requirement, while $(a,b) = (8,8), (9,9)$ do. So, we conclude that the possible values that $b$ can take to satisfy the requirement of the problem are $8, 9, 10, 13, 17, 19, 27$.

Alternate Solution:

Let us denote by $x, y$ and $z, w$ the two numbers that the players $A$ and $B$ wrote down, respectively. We may assume that $x \ge y$ and $z \ge w$ are satisfied. The conditions of the problems can be stated as
$$xy = 2(z+w), \quad zw = 2(x+y), \quad x+y \ge z+w.$$

Adding the respective sides of the first two equations, we obtain $xy + zw = 2(x + y + z + w)$, which can be transformed into the form
$$(x-2)(y-2) + (z-2)(w-2) = 8.$$
First, let us consider the case where $y = 1$. Since $x = 2z + 2w$, we get $zw = 2(x + y) = 4z + 4w + 2$, which is transformed to $(z - 4)(w - 4) = 18$. From this we obtain $(x, w) = (22, 5)$, $(13, 6)$, $(10, 7)$. We find the corresponding values of $x$ are $54, 38, 34$, respectively, and the requirement $x + y \geq z + w$ is also satisfied in all of the three cases. We also find that $z + w = 27, 19, 17$, respectively.
When $w = 1$, we can argue in the same way as above, and find the other two are no solutions satisfying the conditions of the problem.
Next, we assume that $x \geq y \geq 2$ and $z \geq w \geq 2$ are satisfied. Then, we have $(x-2)(y-2) \geq 0$ and $(z-2)(w-2) \geq 0$, and since $(x-2)(y-2)+(z-2)(w-2) = 8$, we must have $(x-2)(y-2) \leq 8$ and $(z-2)(w-2) \leq 8$. If $y \geq 5$, then we get $(x-2)(y-2) \geq (y-2)^2 \geq 9$. So, we must have $y \leq 4$.
If $y = 4$, then from $(x-2)(y-2) \leq 8$, we conclude that $x = 4$ or $5$ or $6$.
* When $x = 4$: From $zw = 2(x + y) = 16$, $z + w = \frac{xy}{2} = 8$, we get $(x, w) = (4, 4)$.
* When $x = 5$: We see there are no pairs $(x, w)$ satisfying $zw = 18$ and $z + w = 10$.
* When $x = 6$: We get $zw = 20$ and $z + w = 12$, which yield $(x, w) = (10, 2)$.
We see that among the possibilities found for $(x, w)$ above only $(x, w) = (4, 4)$ satisfies the condition $x + y \geq z + w$. So, the only solution we get for the case $y = 4$ is $(x, y, z, w) = (4, 4, 4, 4)$ for which $z + w = 8$.
If $y = 3$, then from $xy = 2(z + w)$ we see that $x$ must be even, and from $(x-2)(y-2) \leq 8$, $x$ must be one of $4, 6, 8, 10$.
* When $x = 4$: There are no $(x, w)$ satisfying $zw = 14$, $z + w = 6$.
* When $x = 6$: Only $(x, w) = (6, 3)$ satisfies $zw = 18$, $z + w = 9$.
* When $x = 8$: There are no $(x, w)$ satisfying $zw = 22$, $z + w = 12$.
* When $x = 10$: Only $(x, w) = (13, 2)$ satisfies $zw = 26$, $z + w = 15$.
Among the possibilities found above, we see that only $(x, y, z, w) = (6, 3, 6, 3)$ satisfies the additional condition $x + y \geq z + w$, and we have $z + w = 9$ in this case.
Finally, when $y = 2$, we see that from $(z-2)(w-2) = 8$, we must have $w = 3$ or $w = 4$. When $w = 3$, then $z = 10$ and $x = \frac{2(z+w)}{y} = 13$, and $(x, y, z, w) = (13, 2, 10, 3)$ satisfies the requirements, and we get $z+w = 13$ in this case. If $w = 4$, then we have $z = 6$, $x = \frac{2(z+w)}{y} = 10$, and $(x, y, z, w) = (10, 2, 6, 4)$ satisfies the requirements, and we get $z+w = 10$.
Summarizing what we considered above, we get the possible values for the sum of the two numbers $B$ wrote to be $8, 9, 10, 13, 17, 19, 27$.