Problem 1392

1. We are given 150 numbers $x_1, x_2, \ldots, x_{150}$, each of which is either $\sqrt{2} + 1$ or $\sqrt{2} - 1$.
2. We need to calculate the sum $S = x_1 x_2 + x_3 x_4 + x_5 x_6 + \cdots + x_{149} x_{150}$.
3. Notice that there are 75 terms in the sum $S$, and each term can independently be one of the following:
$$(\sqrt{2} + 1)(\sqrt{2} - 1) = 1, \quad (\sqrt{2} + 1)^2 = 3 + 2\sqrt{2}, \quad (\sqrt{2} - 1)^2 = 3 - 2\sqrt{2}$$
4. Let $x$ be the number of terms equal to $1$, $y$ be the number of terms equal to $3 + 2\sqrt{2}$, and $z$ be the number of terms equal to $3 - 2\sqrt{2}$.
5. We have the following equations:
$$x + y + z = 75$$
$$x + (3 + 2\sqrt{2})y + (3 - 2\sqrt{2})z = S$$
6. Since $S$ is radical-free, the terms involving $\sqrt{2}$ must cancel out. This implies $y = z$.
7. Let $y = z$. Then, we have:
$$x + 2y = 75$$
$$x + 3y = S$$
8. Let $x = 2n + 1$ (since $x$ must be odd). Then:
$$2n + 1 + 2y = 75 \implies n + y = 37 \quad \text{(Equation 1)}$$
$$2n + 1 + 6y = S \implies n + 3y = \frac{S - 1}{2} \quad \text{(Equation 2)}$$
9. For $S = 121$:
$$n + 3y = 60 \quad \text{(Equation 3)}$$
Solving Equations 1 and 3:
$$n + y = 37$$
$$n + 3y = 60$$
Subtracting the first equation from the second:
$$2y = 23 \implies y = 11.5$$
Since $y$ must be an integer, there is no solution for $S = 121$.

10. For $S = 111$:
$$n + 3y = 55 \quad \text{(Equation 4)}$$
Solving Equations 1 and 4:
$$n + y = 37$$
$$n + 3y = 55$$
Subtracting the first equation from the second:
$$2y = 18 \implies y = 9$$
Substituting $y = 9$ into Equation 1:
$$n + 9 = 37 \implies n = 28$$
Therefore, for $S = 111$, we have $n = 28$ and $y = 9$, which is a valid solution.

The final answer is $\boxed{ S = 111 }$ is possible, $S = 121$ is not possible.