Olya wrote fractions of the form $1 / n$ on cards, where $n$ is all possible divisors the numbers $6^{100}$ (including the unit and the number itself). These cards she laid out in some order. After that, she wrote down the number on the first card, then the sum of the numbers on the first and second cards, then the sum of the numbers on the first three cards, etc., finally, the sum of the numbers on all the cards. Every amount Olya recorded on the board in the form of irreducible fraction. What is the least different denominators could be on the numbers on the board?
1. **Identify the divisors of \(6^{100}\)**:
   - The number \(6^{100}\) can be factored as \((2 \cdot 3)^{100} = 2^{100} \cdot 3^{100}\).
   - The divisors of \(6^{100}\) are of the form \(2^a \cdot 3^b\) where \(0 \leq a \leq 100\) and \(0 \leq b \leq 100\).

2. **Rewrite the fractions**:
   - Each divisor \(d\) of \(6^{100}\) can be written as \(\frac{d}{6^{100}}\).
   - Therefore, the fractions are \(\frac{2^a \cdot 3^b}{6^{100}}\).

3. **Sum of the fractions**:
   - The sum of all these fractions is:
     \[
     \sum_{a=0}^{100} \sum_{b=0}^{100} \frac{2^a \cdot 3^b}{6^{100}} = \sum_{a=0}^{100} \sum_{b=0}^{100} \frac{2^a \cdot 3^b}{2^{100} \cdot 3^{100}} = \sum_{a=0}^{100} \sum_{b=0}^{100} \frac{1}{2^{100-a} \cdot 3^{100-b}}
     \]

4. **Simplify the sum**:
   - Notice that each term \(\frac{1}{2^{100-a} \cdot 3^{100-b}}\) is a fraction with a denominator that is a divisor of \(6^{100}\).
   - The sum of all these fractions is:
     \[
     \sum_{a=0}^{100} \sum_{b=0}^{100} \frac{1}{2^{100-a} \cdot 3^{100-b}} = \sum_{a=0}^{100} \sum_{b=0}^{100} \frac{1}{6^{100}} = \frac{1}{6^{100}} \sum_{a=0}^{100} \sum_{b=0}^{100} 1 = \frac{1}{6^{100}} \cdot 101 \cdot 101 = \frac{101^2}{6^{100}}
     \]

5. **Determine the least number of different denominators**:
   - The sums written on the board are cumulative sums of the fractions.
   - The first sum is \(\frac{1}{6^{100}}\), the second sum is \(\frac{1}{6^{100}} + \frac{1}{6^{100}} = \frac{2}{6^{100}}\), and so on.
   - The denominators of these sums are \(6^{100}\) and its multiples.

6. **Identify the distinct denominators**:
   - The denominators of the sums can be either \(6^{100}\) or \(\frac{6^{100}}{2}\) (since the sums are multiples of \(\frac{1}{6^{100}}\)).
   - Therefore, the least number of different denominators is 2.

\(\blacksquare\)

The final answer is \( \boxed{ 2 } \).