You are given $n$ not necessarily distinct real numbers $a_1, a_2, \ldots, a_n$. Let's consider all $2^n-1$ ways to select some nonempty subset of these numbers, and for each such subset calculate the sum of the selected numbers. What largest possible number of them could have been equal to $1$?

For example, if $a = [-1, 2, 2]$, then we got $3$ once, $4$ once, $2$ twice, $-1$ once, $1$ twice, so the total number of ones here is $2$.

[i](Proposed by Anton Trygub)[/i]
1. **Understanding the Problem:**
   We are given \( n \) real numbers \( a_1, a_2, \ldots, a_n \). We need to find the largest possible number of nonempty subsets of these numbers such that the sum of the elements in each subset is equal to 1.

2. **Analyzing Subsets:**
   There are \( 2^n \) subsets of \( n \) elements, including the empty subset. Since we are only interested in nonempty subsets, we have \( 2^n - 1 \) nonempty subsets.

3. **Considering the Sum of Subsets:**
   If all elements \( a_i \) are zero, then the sum of any subset is zero. Therefore, no subset can have a sum of 1 in this case.

4. **Introducing a Nonzero Element:**
   Suppose at least one element \( x_0 \) is nonzero. We need to consider subsets that include \( x_0 \) and those that do not.

5. **Subsets Including \( x_0 \):**
   If a subset \( S \) with sum 1 contains \( x_0 \), then the sum of the subset \( S \setminus \{x_0\} \) (the subset without \( x_0 \)) must be \( 1 - x_0 \).

6. **Subsets Excluding \( x_0 \):**
   If a subset \( S \) with sum 1 does not contain \( x_0 \), then adding \( x_0 \) to this subset will result in a sum \( \ne 1 \).

7. **Counting Subsets:**
   The number of subsets with sum 1 is at most half of the total number of nonempty subsets. This is because for each subset with sum 1, there is a corresponding subset (by adding or removing \( x_0 \)) that does not sum to 1.

8. **Maximum Number of Subsets with Sum 1:**
   Therefore, the maximum number of subsets with sum 1 is \( \le \frac{2^n - 1}{2} \). Simplifying, we get:
   \[
   \frac{2^n - 1}{2} = 2^{n-1} - \frac{1}{2}
   \]
   Since the number of subsets must be an integer, we take the floor function:
   \[
   \left\lfloor 2^{n-1} - \frac{1}{2} \right\rfloor = 2^{n-1}
   \]

9. **Achieving the Maximum:**
   To achieve this maximum, consider the set \( \{1, 0, 0, \ldots, 0\} \). Here, the number of subsets with sum 1 is exactly \( 2^{n-1} \), as each subset containing the element 1 and any combination of zeros will sum to 1.

Conclusion:
\[
\boxed{2^{n-1}}
\]