Let $n=2^{31}3^{19}.$  How many positive integer divisors of $n^2$ are less than $n$ but do not divide $n$?
1. **Determine the prime factorization of \( n \):**
   \[
   n = 2^{31} \cdot 3^{19}
   \]

2. **Calculate \( n^2 \):**
   \[
   n^2 = (2^{31} \cdot 3^{19})^2 = 2^{62} \cdot 3^{38}
   \]

3. **Find the total number of divisors of \( n^2 \):**
   The number of divisors of a number \( a^m \cdot b^n \) is given by \((m+1)(n+1)\). Therefore, the number of divisors of \( n^2 \) is:
   \[
   (62+1)(38+1) = 63 \cdot 39 = 2457
   \]

4. **Determine the number of divisors of \( n^2 \) that are less than \( n \):**
   Since \( n^2 \) is a perfect square, its divisors can be paired such that the product of each pair is \( n^2 \). For example, if \( d \) is a divisor of \( n^2 \), then \( \frac{n^2}{d} \) is also a divisor. One of these divisors in each pair will be less than \( n \) and the other will be greater than \( n \). Since \( n^2 \) has 2457 divisors, and one of them is \( n \) itself, the number of divisors less than \( n \) is:
   \[
   \frac{2457 - 1}{2} = 1228
   \]

5. **Find the number of divisors of \( n \):**
   The number of divisors of \( n = 2^{31} \cdot 3^{19} \) is:
   \[
   (31+1)(19+1) = 32 \cdot 20 = 640
   \]
   Note that this includes \( n \) itself, so the number of divisors of \( n \) other than \( n \) is:
   \[
   640 - 1 = 639
   \]

6. **Calculate the number of divisors of \( n^2 \) that are less than \( n \) but do not divide \( n \):**
   Subtract the number of divisors of \( n \) (other than \( n \) itself) from the number of divisors of \( n^2 \) that are less than \( n \):
   \[
   1228 - 639 = 589
   \]

The final answer is \(\boxed{589}\)