Solution:
The answer is all $n \neq 2$. For $n=1$, the set $\{1\}$ works. For $n=2$, no set exists, simply because the sum of reciprocals of two distinct integers cannot be equal to $1$. For $n=3$, take $\{2,3,6\}$.

For $n>3$, the identity
$$\frac{1}{k} = \frac{1}{k+r} + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} + \cdots + \frac{1}{(k+r-1)(k+r)}$$
allows us to extend a sum of $t$ terms to one of exactly $t+r$ terms. Taking $k=3$ and $r=n-3$ allows us to turn the sum $1 = 1/2 + 1/3 + 1/6$ to the $n$-term sum
$$1 = \frac{1}{2} + \frac{1}{6} + \frac{1}{n} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \cdots + \frac{1}{(n-1)n}$$
This construction works provided that $n \neq k(k+1)$ for any $k$. Otherwise, we have $n \geq 6$, and instead we apply the above to the sum $1 = 1/2 + 1/3 + 1/10 + 1/15$, taking $k=3$, $r=n-4$ to yield
$$1 = \frac{1}{2} + \frac{1}{10} + \frac{1}{15} + \frac{1}{n-1} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \cdots + \frac{1}{(n-2)(n-1)}$$
The above then works, because if $n = k(k+1)$ for some $k$, then $n-1 \neq 2, 10, 15$, and $n-1 \neq m(m+1)$ for any $m$ by parity, since $n-1$ is odd and $m(m+1)$ is always even. This construction is not unique; there are other similar ones.