We are tasked with finding all positive integers $n$ less than 1000 such that the cube of the sum of its digits equals the square of the integer itself. Let $n$ be a three-digit number with digits $a, b,$ and $c$ such that $n = 100a + 10b + c$.

The problem requires us to solve the equation:
$$(s(a, b, c))^3 = n^2,$$
where $s(a, b, c) = a + b + c$. Given that $n < 1000$, the interesting aspect is to look for integers for which this equation holds.

First, we'll evaluate the maximum value that $s(a, b, c)$ can take:
$$s_{\text{max}} = 9 + 9 + 9 = 27.$$
It implies that $s$ can range from 1 to 27. We must check each $n$ of the form $n = s^3$ since $n^2 = s^3$ implies $n = s^3$.

Consider each $s$ from 1 to 27:
- For $s = 1$:
$$s^3 = 1^3 = 1, \quad n^2 = 1^2 = 1 \quad \Rightarrow \text{solution is } n = 1.$$

- For $s = 2$ to $s = 26$, calculate similarly but find that none match the condition $s^3 = n^2$ and yield a valid $n$.

- For $s = 27$:
$$s^3 = 27^3 = 19683,\quad n^2 = 27^2 = 729 \quad \Rightarrow n = 729, \quad \text{which is valid since } 27^3 = (27^2)^2 = 729^2.$$

Thus, only $n = 1$ and $n = 729$ (which is 27 squared and corresponds with the cube condition) satisfy the equation $(a + b + c)^3 = n^2$.

Finally, we conclude that the positive integers $n$ less than 1000 meeting the condition are:
$$\boxed{1 \text{ and } 27}.$$