How many integers $n$ are there subject to the constraint that $1 \leq n \leq 2020$ and $n^n$ is a perfect square? 
1. **Case 1: \( n \) is even**

   If \( n \) is even, then \( n = 2k \) for some integer \( k \). We need to check if \( n^n \) is a perfect square.
   
   \[
   n^n = (2k)^{2k}
   \]
   
   Since \( 2k \) is even, \( (2k)^{2k} \) is always a perfect square because any even number raised to an even power is a perfect square. 

   The even numbers between 1 and 2020 are:
   
   \[
   2, 4, 6, \ldots, 2020
   \]
   
   This is an arithmetic sequence with the first term \( a = 2 \) and common difference \( d = 2 \). The number of terms in this sequence is:
   
   \[
   n = \frac{2020 - 2}{2} + 1 = 1010
   \]
   
   So, there are 1010 even integers \( n \) such that \( n^n \) is a perfect square.

2. **Case 2: \( n \) is odd**

   If \( n \) is odd, then \( n \) must be a perfect square for \( n^n \) to be a perfect square. This is because if \( n \) is odd and not a perfect square, \( n^n \) will not be a perfect square.
   
   The odd perfect squares between 1 and 2020 are:
   
   \[
   1^2, 3^2, 5^2, \ldots, 43^2
   \]
   
   This is an arithmetic sequence with the first term \( a = 1 \) and common difference \( d = 2 \). The largest odd perfect square less than or equal to 2020 is \( 43^2 = 1849 \). The number of terms in this sequence is:
   
   \[
   n = \frac{43 - 1}{2} + 1 = 22
   \]
   
   So, there are 22 odd integers \( n \) such that \( n^n \) is a perfect square.

3. **Total number of integers \( n \) such that \( n^n \) is a perfect square**

   Adding the results from both cases:
   
   \[
   1010 + 22 = 1032
   \]

Conclusion:
\[
\boxed{1032}
\]