Iranian Mathematical Olympiad

We claim that the answer is $n-2$. First, note that since the sum of a number of variables (at least 2) is one, and the rest of the numbers are less than one, the sum of all the variables is less than $n-1$. Suppose the numbers are $a_i = 1-b_i$.

First solution. Using induction, we will prove that for each $n$, there exists $b_1, \dots, b_n$ satisfying the following properties:
$$\bullet \quad \forall 1 < k < n-1, \exists i \le k \le n : b_{i_1} + \dots + b_{i_k} = 1$$
$$\bullet \quad b_1 + \dots + b_n = 2$$
The given conditions imply the desired result because:
$$\forall 1 < k < n-1 : a_{i_1} + \dots + a_{i_k} = k-1 \in \{1, \dots, n-3\}$$
And also:
$$a_1 + \dots + a_n = n - 2$$

In other words, we divide all previous terms by 2 and add two new $\frac{1}{2}$. Now we check the correctness of the induction hypothesis:

1.
$$k = 2 : b'_{n+1} + b'_{n+2} = 1$$

2.
$$3 \le k < n : \exists b_{i_1} + \cdots + b_{i_{k-1}} = 1 \implies b_1' + \cdots + b_{i_{k-1}}' + b_{n+1} = 1$$

3.
$$k = n : b_1' + \cdots + b_n' = 1$$

4.
$$b_1' + \cdots + b_{n+2}' = 1 + \frac{b_1 + \cdots + b_n}{2} = 2$$

For the base of induction we use:
$$n = 3 : \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \quad n = 4 : \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}$$

Second Solution
$b_{i_1} + \cdots + b_{i_t} = 1$ implies $a_{i_1} + \cdots + a_{i_t} = t - 1$. Thus, we should give an example that for $2 \le t \le n - 1$ sum of $t$ of them is 1.
Let $\alpha = \frac{\sqrt{5}-1}{2}$ which is the root of $x^2+x=1$ and $0 < \alpha < 1$. If $n=2m$ is even, take $b_k = a^k$ for $k=1, \cdots, 2m-2$ and $b_{2m-1} = b_{2m} = \frac{1}{2}\alpha^{2m-3}$. Now for all $1 \le t \le m-1$ it follows that
$$b_{2t}+b_{2t-1}+b_{2t-3}+b_{2t-5}+\cdots+b_1 = b_{2t-2}+b_{2t-3}+b_{2t-5}+\cdots+b_1 = \cdots = b_2+b_1 = 1$$
Thus the sum of $t+1$ of $b_i$'s is 1.
Furthermore, since $\alpha^{2m-2}+\alpha^{2m-3}+\cdots+\alpha^{2t} = \frac{\alpha^{2t}-\alpha^{2m-1}}{1-\alpha} = \frac{\alpha^{2t}-\alpha^{2m-1}}{\alpha^2} = \alpha^{2t-2}-\alpha^{2m-3}$ we have
$$b_{2m} + b_{2m-1} + b_{2m-2} + \cdots + b_2 + b_{2t-3} + b_{2t-5} + \cdots + b_1 = 1$$
Hence sum of $2m-t$ of $b_i$'s is 1. If $n=2m-1$ is odd, then setting $b_k = \alpha^k$ for $k=1, \cdots, 2m-2$ and $b_{2m-1} = \alpha^{2m-3}$, we have
$$b_{2t}+b_{2t-1}+b_{2t-3}+\cdots+b_1 = b_{2m-1}+b_{2m-2}+\cdots+b_2+b_{2t-3}+b_{2t-5}+\cdots+b_1 = 1$$
We proved that for all $2 \le t \le 2m-2$ the sum of $t$ of $b_i$'s is 1. ■