Observe that $1183 = 13^2 \cdot 7$, so $\sqrt{1183} = 13\sqrt{7}$. Because $\sqrt{x} + \sqrt{y} = 13\sqrt{7}$, it follows that $\sqrt{x}$ and $\sqrt{y}$ must be of the form $a\sqrt{7}$ and $b\sqrt{7}$, respectively, where $a$ and $b$ are positive integers and $a + b = 13$. Then $\sqrt{x} = \sqrt{7a^2}$ and $\sqrt{y} = \sqrt{7b^2}$, so $x = 7a^2$ and $y = 7b^2$. Substituting gives
$$x + y = 7(a^2 + b^2) = 7(a^2 + (13-a)^2) = 14a^2 - 182a + 1183.$$
The minimum value of the quadratic polynomial occurs at $a = \frac{182}{2 \cdot 14} = 6.5$. Because $a$ and $b$ must be positive integers, without loss of generality (by symmetry), choose $a = 6$ and $b = 7$. The minimum possible value of $x + y$ is
$$x + y = 7 \cdot (6^2 + 7^2) = 7 \cdot (36 + 49) = 7 \cdot 85 = 595.$$