The answer is that the given property holds for all $n \ne 2$. For $n = 1$, the set $\{1\}$ satisfies (3). For $n = 2$, note that no set satisfies (3); if $a_1$ or $a_2$ equals $1$, then $\frac{1}{a_1} + \frac{1}{a_2} > 1$, if $a_1$ and $a_2$ are both at least two, then $\frac{1}{a_1} + \frac{1}{a_2} \le \frac{1}{2} + \frac{1}{3} < 1$.

$$\frac{1}{k} = \frac{1}{k+\ell} + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} + \cdots + \frac{1}{(k+\ell-1)(k+\ell)}. \quad (4)$$
Indeed, if we use $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$ then the right-hand side of (4) is equal to
$$\frac{1}{k+\ell} + \left(\frac{1}{k} - \frac{1}{k+1}\right) + \left(\frac{1}{k+1} - \frac{1}{k+2}\right) + \cdots + \left(\frac{1}{k+\ell-1} - \frac{1}{k+\ell}\right) = \frac{1}{k+\ell} + \frac{1}{k} - \frac{1}{k+\ell}.$$

Substituting $k=1$ and $\ell=n-1$ into (4), we obtain an identity
$$1 = \frac{1}{n} + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \cdots + \frac{1}{(n-1) \cdot n}. \qquad (5)$$
If $n \neq k(k+1)$ for all $k \ge 1$, this is a sum of $n$ distinct reciprocals, each with denominator smaller than $n^2$. This shows that the given property holds for all $n \ge 3$ not of the form $k(k+1)$.

Suppose that there exists some $k \ge 1$ such that $n = k(k+1)$. Then we apply to the right-hand side of (5) the substitutions $\frac{1}{n} + \frac{1}{(n-1)n} = \frac{1}{n-1}$ and $\frac{1}{6} = \frac{1}{10} + \frac{1}{15}$. Then we get the identity
$$1 = \frac{1}{n-1} + \frac{1}{2} + \frac{1}{10} + \frac{1}{15} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \cdots + \frac{1}{(n-2) \cdot (n-1)}. \quad (6)$$
This is a sum of $n$ reciprocals. Note that each denominator of each reciprocal is smaller than $n^2$; this is only non-obvious for the term $\frac{1}{15}$, and since $n = k(k+1)$ and $n \ge 3$, we in particular have $n \ge 6$, and therefore $n^2 > 15$. Now we show that these reciprocals are distinct.
Note that $k(k+1)$ is always even. Since $n$ is of the form $k(k+1)$, $n-1$ is odd and therefore not of this form. Furthermore, $n-1$ is also unequal to $2$, $10$ and $15$, since $3$, $11$ and $16$ are not of the form $k(k+1)$. Also, $10$ and $15$ themselves are not of the form $k(k+1)$. Therefore all the reciprocals in the right-hand side of (6) are distinct. So the given property holds for all $n \ge 3$ of the form $k(k+1)$ as well. $\square$