The sum of two real numbers is a positive integer $$𝑛$$ and the sum of their squares is $$𝑛 + 1012$$. Find the maximum possible value of $$n$$.

Let the two real numbers be $$x$$ and $$y$$.

Their sum is given to be a positive integer $$n$$, so
$$x + y = n \qquad -(1)$$

The sum of their squares is
$$x^{2} + y^{2} = n + 1012 \qquad -(2)$$

For any two real numbers with a fixed sum, the expression $$x^{2}+y^{2}$$ attains its minimum value when the numbers are equal. Using the identity
$$x^{2}+y^{2} = (x+y)^{2} - 2xy,$$
and the AM-GM inequality $$xy \le \left(\frac{x+y}{2}\right)^{2},$$ we obtain

Minimum value of $$x^{2}+y^{2}$$ = $$\dfrac{(x+y)^{2}}{2} = \dfrac{n^{2}}{2}$$ when $$x = y = \dfrac{n}{2}$$.

Because $$x^{2}+y^{2}$$ in our problem equals $$n+1012,$$ it must be at least this minimum value:
$$n+1012 \;\ge\; \dfrac{n^{2}}{2} \qquad -(3)$$

Re-arrange inequality (3):
$$\dfrac{n^{2}}{2} - n - 1012 \le 0$$
$$n^{2} - 2n - 2024 \le 0 \qquad -(4)$$

Solve the quadratic equality $$n^{2}-2n-2024 = 0$$:
Discriminant $$\Delta = (-2)^{2} - 4(1)(-2024) = 4 + 8096 = 8100$$
$$\sqrt{\Delta} = 90$$

Roots are
$$n = \dfrac{2 \pm 90}{2} \;=\; 46 \text{ or } -44$$

Inequality (4) is satisfied for $$-44 \le n \le 46$$. Since $$n$$ is required to be a positive integer, the largest admissible value is $$n = 46$$.

Check attainability: take $$x = y = \dfrac{n}{2} = 23$$.
Then $$x+y = 46$$ and $$x^{2}+y^{2} = 23^{2}+23^{2} = 2 \times 529 = 1058 = 46 + 1012$$, satisfying both conditions.

Hence the maximum possible value of $$n$$ is 46.

Final Answer: 46