Solution:
We have $1^{3} + 2^{3} + \cdots + 16^{3} = (1 + 2 + \cdots + 16)^{2} = 136^{2}$.

So, $1^{3} + 2^{3} + \cdots + 16^{3} + 17^{n} = 136^{2} + 17^{n}$.

We want $136^{2} + 17^{n}$ to be a perfect square.

Let $136^{2} + 17^{n} = k^{2}$ for some integer $k$.

Then $k^{2} - 136^{2} = 17^{n}$, so $(k - 136)(k + 136) = 17^{n}$.

Since $17$ is prime, both $k - 136$ and $k + 136$ must be powers of $17$.

Let $k - 136 = 17^{a}$, $k + 136 = 17^{b}$, with $a < b$, $a + b = n$.

Then $k + 136 - (k - 136) = 272 = 17^{b} - 17^{a}$.

So $272 = 17^{a}(17^{b - a} - 1)$.

Now, $17^{a}$ divides $272$. Since $272 = 16 \times 17$, the possible values for $a$ are $0$ and $1$.

Case 1: $a = 0$

Then $17^{b - 0} - 1 = 272$, so $17^{b} = 273$, which is not possible.

Case 2: $a = 1$

Then $17^{1}(17^{b - 1} - 1) = 272$, so $17^{b - 1} - 1 = 16$, so $17^{b - 1} = 17$, so $b - 1 = 1$, $b = 2$.

So $a = 1$, $b = 2$, $n = a + b = 3$.

Therefore, the only solution is $n = 3$.

Check:
$1^{3} + 2^{3} + \cdots + 16^{3} + 17^{3} = 136^{2} + 4913 = 18496 + 4913 = 23409 = 153^{2}$, which is a perfect square.

Thus, the only positive integer $n$ for which $1^{3} + 2^{3} + \cdots + 16^{3} + 17^{n}$ is a perfect square is $n = 3$.