To solve this problem, let’s analyze the equation:

$\sqrt{x} + \sqrt{y} = \sqrt{1183}$

We want to minimize the sum $x + y$, where $x$ and $y$ are positive integers. Here's how we can proceed:

1. Rewrite the Equation: Let $\sqrt{x} = a$ and $\sqrt{y} = b$, where $a$ and $b$ are real numbers such that $a + b = \sqrt{1183} \approx 34.39$.

2. Minimize $x + y$: Since $x = a^2$ and $y = b^2$, our goal is to find values of $a$ and $b$ that add up to approximately 34.39 while keeping $x = a^2$ and $y = b^2$ as integers and minimizing $x + y$.

3. Finding Integer Values:

   • We start by testing values close to half of $\sqrt{1183}$, which is approximately 17.2.
   • Let's try $a = 17$ and $b = 34.39 - 17 = 17.39$, then square these values to check if they yield integers for $x$ and $y$.

4. Calculate $x$ and $y$ for $a = 17$ and $b = 18$:

   • If $a = 17$, then $x = 17^2 = 289$.
   • If $b = 18$, then $y = 18^2 = 324$.

5. Verify the Solution:

   • $\sqrt{x} + \sqrt{y} = 17 + 18 = 35$, which is approximately $\sqrt{1183}$.
   • The sum $x + y = 289 + 324 = 613$.

Thus, the least possible sum of $x + y$ is:

$\boxed{613}$

Would you like further details, or do you have any questions?

Here are some related questions for further exploration:

1. What happens if we slightly increase or decrease $a$ and $b$?
2. How does the result change if $\sqrt{1183}$ were a different number?
3. Can we find other pairs $(x, y)$ that satisfy the same condition with different values?
4. How would the approach change if $x$ and $y$ were required to be consecutive integers?
5. What if the equation involved cube roots instead of square roots?

Tip: When trying to minimize the sum of squares, working with values close to the square root of half the target number often yields optimal solutions.