IMO Team Selection Test 1

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 $\frac{1}{a_1}+\frac{1}{a_2}\le \frac{1}{2}+\frac{1}{3}<1$.

$$\frac{1}{k}=\frac{1}{k+\ell }+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\cdots +\frac{1}{(k+\ell -1)(k+\ell )}.(4)$$ 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 $$1=\frac{1}{n}+\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\cdots +\frac{1}{(n-1)\cdot n}.(5)$$ If $n\ne k(k+1)$ 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 $$1=\frac{1}{n-1}+\frac{1}{2}+\frac{1}{10}+\frac{1}{15}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\cdots +\frac{1}{(n-2)\cdot (n-1)}.(6)$$ 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. $\square$