Solution:
Let $n = 2^{23} 3^{17}$.

The number of factors of $n$ is $(23+1)(17+1) = 24 \times 18 = 432$.

The number of factors of $n^2$ is $(2 \times 23 + 1)(2 \times 17 + 1) = 47 \times 35 = 1645$.

The factors of $n^2$ come in pairs $(d, n^2/d)$, and exactly one of each pair is less than $n$ (unless $d = n$). Since $n^2$ is not a perfect square (because $n$ is not a perfect square), $n$ is not a factor of $n^2$ such that $n^2 = n \times n$ with $n$ integer, but in this case $n$ is a factor of $n^2$.

But $n^2$ is a perfect square, and $n$ is a factor of $n^2$, and $n^2 = n \times n$.

So, the number of factors of $n^2$ less than $n$ is $\frac{1645 - 1}{2} = 822$.

Now, among these, how many do not divide $n$?

The factors of $n$ are among the factors of $n^2$, and all factors of $n$ are less than or equal to $n$ (except $n$ itself). So, among the $432$ factors of $n$, $n$ itself is counted, and $431$ are less than $n$.

Therefore, the number of factors of $n^2$ less than $n$ that do not divide $n$ is $822 - 431 = 391$.

Answer: $391$.