For positive integers $n,$ let $\nu_3 (n)$ denote the largest integer $k$ such that $3^k$ divides $n.$ Find the number of subsets $S$ (possibly containing 0 or 1 elements) of $\{1, 2, \ldots, 81\}$ such that for any distinct $a,b \in S$, $\nu_3 (a-b)$ is even.

[i]Author: Alex Zhu[/i]

[hide="Clarification"]We only need $\nu_3(a-b)$ to be even for $a>b$. [/hide]
To solve the problem, we need to find the number of subsets \( S \) of \(\{1, 2, \ldots, 81\}\) such that for any distinct \( a, b \in S \), \(\nu_3(a - b)\) is even. 

1. **Understanding \(\nu_3(a - b)\):**
   \(\nu_3(n)\) denotes the largest integer \( k \) such that \( 3^k \) divides \( n \). For \(\nu_3(a - b)\) to be even, \( a \) and \( b \) must be congruent modulo \( 3^k \) for some even \( k \).

2. **Classifying numbers modulo 3:**
   Consider the numbers modulo 3. The set \(\{1, 2, \ldots, 81\}\) can be divided into three residue classes modulo 3: 
   \[
   \{1, 4, 7, \ldots, 79\}, \{2, 5, 8, \ldots, 80\}, \{3, 6, 9, \ldots, 81\}
   \]
   Each class contains 27 numbers.

3. **Ensuring \(\nu_3(a - b)\) is even:**
   For \(\nu_3(a - b)\) to be even, \( a \) and \( b \) must be congruent modulo 9. This is because if \( a \equiv b \pmod{9} \), then \( a - b \) is divisible by 9, and \(\nu_3(a - b) \geq 2\).

4. **Classifying numbers modulo 9:**
   Each residue class modulo 3 can be further divided into three residue classes modulo 9. For example, the class \(\{1, 4, 7, \ldots, 79\}\) can be divided into:
   \[
   \{1, 10, 19, \ldots, 73\}, \{4, 13, 22, \ldots, 76\}, \{7, 16, 25, \ldots, 79\}
   \]
   Each of these classes contains 9 numbers.

5. **Counting subsets:**
   For each residue class modulo 9, we can choose any subset of the numbers in that class. There are \(2^9\) ways to choose a subset from 9 elements (including the empty set).

6. **Combining subsets:**
   Since there are 9 residue classes modulo 9 (3 for each of the 3 residue classes modulo 3), and we can choose independently from each class, the total number of valid subsets is:
   \[
   (2^9)^9 = 2^{81}
   \]

The final answer is \(\boxed{2^{81}}\)