Problem 1782

Each of the sums $a_1 + 2a_2 + \dots + n a_n$ and $\frac{1}{a_1} + \frac{2}{a_2} + \dots + \frac{n}{a_n}$ contains $1 + 2 + \dots + n = \frac{n(n+1)}{2}$ numbers. From the given data, the arithmetic mean $A$ and the harmonic mean $H$ of numbers $a_1, a_2, a_3, a_3, a_3, \dots, a_n, a_n, \dots, a_n$ can be calculated. We get
$$A = \frac{a_1 + 2a_2 + \dots + n a_n}{\frac{n(n+1)}{2}} = \frac{6n}{\frac{n(n+1)}{2}} = \frac{12}{n+1}$$
and
$$H = \frac{\frac{n(n+1)}{2}}{\frac{1}{a_1} + \frac{2}{a_2} + \dots + \frac{n}{a_n}} = \frac{\frac{n(n+1)}{2}}{2 + \frac{1}{n}} = \frac{n^2(n+1)}{2(2n+1)}$$
Since all the numbers are positive, the inequality of arithmetic and geometric means states
$$\frac{n^2(n+1)}{2(2n+1)} = H \le A = \frac{12}{n+1},$$
which is equivalent to
$$n^2(n+1)^2 \le 24(2n+1).$$
But for $n \ge 4$ we have $\frac{n^3}{2} \ge \frac{64}{2} = 32 > 24$ and hence
$$n^2(n+1)^2 = \frac{n^2(n+1)(2n+2)}{2} \ge (2n+1)\frac{n^3}{2} > 24(2n+1),$$
from which we conclude that $n$ may be at most $3$.

For $n=1$ we get the conditions $a_1 = 6$ and $\frac{1}{a_1} = 3$, which is impossible. For $n=2$ we get the conditions $a_1 + 2a_2 = 12$ and $\frac{1}{a_1} + \frac{2}{a_2} = \frac{5}{2}$. The second condition can now only be satisfied if one of the numbers is equal to $1$. If this is not the case, when $a_1 \ge 2$ and $a_2 \ge 2$, the left side of the equality is too small. The first condition says that only $a_2$ can be odd, hence $a_2 = 1$ and $a_1 = 10$. This is inconsistent with the second condition. We conclude that for $n=2$ again such numbers $a_1$ and $a_2$ do not exist.

In contrast, for $n=3$ such three numbers can be found. Consider $a_1 = 6$, $a_2 = 3$ and $a_3 = 2$. It holds true
$$a_1 + 2a_2 + 3a_3 = 6 + 2 \cdot 3 + 3 \cdot 2 = 18 = 6 \cdot 3$$
and
$$\frac{1}{a_1} + \frac{2}{a_2} + \frac{3}{a_3} = \frac{1}{6} + \frac{2}{3} + \frac{3}{2} = \frac{1+4+9}{6} = \frac{14}{6} = 2 + \frac{1}{3}.$$
The only natural number that solves the problem is thus $3$.