International Mathematical Olympiad

Let $M = \max\{a_1, \dots, a_n\}$. Then we have
$$3^M = 1 \cdot 3^{M-a_1} + 2 \cdot 3^{M-a_2} + \dots + n \cdot 3^{M-a_n} \equiv 1 + 2 + \dots + n = \frac{n(n+1)}{2} \pmod n.$$
Therefore, the number $\frac{n(n+1)}{2}$ must be odd and hence $n \equiv 1 \pmod 4$ or $n \equiv 2 \pmod 4$.

We will now prove that each $n \in \mathbb{N}$ of the form $4k + 1$ or $4k + 2$ (for some $k \in \mathbb{N}$) there exist integers $a_1, \dots, a_n$ with the described property.

For a sequence $\mathbf{a} = (a_1, a_2, \dots, a_n)$ let us introduce the following notation:
$$L(\mathbf{a}) = \frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \dots + \frac{1}{2^{a_n}} \quad \text{and} \quad R(\mathbf{a}) = \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \dots + \frac{n}{3^{a_n}}.$$
Assume that for $n = 2m + 1$ there exists a sequence $\mathbf{a} = (a_1, \dots, a_n)$ of non-negative integers with $L(\mathbf{a}) = R(\mathbf{a}) = 1$. Consider the sequence $\mathbf{a}' = (a'_1, \dots, a'_{n+1})$ defined in the following way:
$$a'_j = \begin{cases} a_j, & \text{if } j \notin \{m+1, 2m+2\} \\ a_{m+1} + 1, & \text{if } j \in \{m+1, 2m+2\}. \end{cases}$$
Then we have
$$L(\mathbf{a}') = L(\mathbf{a}) - \frac{1}{2^{a_{m+1}}} + 2 \cdot \frac{1}{2^{a_{m+1}+1}} = 1$$
$$R(\mathbf{a}') = R(\mathbf{a}) - \frac{m+1}{3^{a_{m+1}}} + \frac{m+1}{3^{a_{m+1}+1}} + \frac{2m+2}{3^{a_{m+1}+1}} = 1.$$
This implies that if the statement holds for $2m + 1$, then it holds for $2m + 2$.

Assume now that the statement holds for $n = 4m + 2$ for some $m \ge 2$, and assume that $\mathbf{a} = (a_1, \dots, a_{4m+2})$ is the corresponding sequence of $n$ non-negative integers. We will construct a following sequence $\mathbf{a}' = (a'_1, a'_2, \dots, a'_{4m+13})$ that satisfies $L(\mathbf{a}') = R(\mathbf{a}') = 1$ thus proving that the statement holds for $4m + 13$. Define:
$$a'_j = \begin{cases} a_{m+2} + 2 & \text{if } j = m + 2 \\ a_j + 1 & \text{if } j \in \{2m + 2, 2m + 3, 2m + 4, 2m + 5, 2m + 6\} \\ a_{j/2} + 1 & \text{if } j \in \{4m + 4, 4m + 6, 4m + 8, 4m + 10, 4m + 12\} \\ a_{m+2} + 3 & \text{if } j \in \{4m + 3, 4m + 5, 4m + 7, 4m + 9, 4m + 11, 4m + 13\} \\ a_j & \text{otherwise.} \end{cases}$$
We now have
$$L(\mathbf{a}') = L(\mathbf{a}) - \frac{1}{2^{a_{m+2}}} - \sum_{j=2}^{6} \frac{1}{2^{a_{2m+j}}} + \frac{1}{2^{a_{m+2}+2}} + \sum_{j=2}^{6} \frac{1}{2^{a_{2m+j}+1}} + \sum_{j=2}^{6} \frac{1}{2^{a_{2m+j}+1}} + 6 \cdot \frac{1}{2^{a_{m+2}+3}} = 1.$$
It remains to verify that $R(\mathbf{a}') = R(\mathbf{a}) = 1$. We write
$$R(\mathbf{a}') - R(\mathbf{a}) = R\left(\begin{array}{ccccccc} a'_{m+2} & a'_{4m+3} & a'_{4m+5} & a'_{4m+7} & a'_{4m+9} & a'_{4m+11} & a'_{4m+13} \\ m+2 & 4m+3 & 4m+5 & 4m+7 & 4m+9 & 4m+11 & 4m+13 \end{array}\right) - R\left(\begin{array}{c} a_{m+2} \\ m+2 \end{array}\right) + \sum_{j=2}^{6} \left(R\left(\begin{array}{c} a'_{2m+j} \\ 2m+j, 4m+j \end{array}\right) - R\left(\begin{array}{c} a_{2m+j} \\ 2m+j \end{array}\right)\right),$$
where
$$R\left(\begin{array}{ccc} c_1 & \dots & c_k \\ d_1 & \dots & d_k \end{array}\right)=\frac{d_1}{3^{c_1}}+\dots+\frac{d_k}{3^{c_k}}.$$
For each $j \in \{2, 3, 4, 5, 6\}$ we have
$$R\left(\begin{array}{cc} a'_{2m+j} & a'_{4m+2j} \\ 2m+j, & 4m+j \end{array}\right)-R\left(\begin{array}{c} a_{2m+j} \\ 2m+j \end{array}\right)=\frac{2m+j}{3^{a_{2m+j}+1}}+\frac{4m+2j}{3^{a_{2m+j}+1}}-\frac{2m+j}{3^{a_{2m+j}}} = 0.$$
The first term in the expression for $R(\mathbf{a}') - R(\mathbf{a})$ is also equal to 0 because
$$\begin{aligned} & R \begin{pmatrix} a'_{m+2}, & a'_{4m+3}, & a'_{4m+5}, & a'_{4m+7}, & a'_{4m+9}, & a'_{4m+11}, & a'_{4m+13} \\ m+2, & 4m+3, & 4m+5, & 4m+7, & 4m+9, & 4m+11, & 4m+13 \end{pmatrix} \ & - R \begin{pmatrix} a_{m+2} \\ m+2 \end{pmatrix} \ &= \frac{m+2}{3^{a_{m+2}+2}} + \sum_{j=1}^{6} \frac{4m+2j+1}{3^{a_{m+2}+3}} - \frac{m+2}{3^{a_{m+2}}} \ &= 0. \end{aligned}$$
Thus $R(\mathbf{a}') = 0$ and the statement holds for $4m + 13$. It remains to verify that there are sequences of lengths 1, 5, 9, 13, and 17. One way to choose these sequences is:
$$(1), \quad (2, 1, 3, 4, 4), \quad (2, 3, 3, 3, 3, 4, 4, 4, 4), \quad (2, 3, 3, 4, 4, 4, 5, 4, 4, 5, 4, 5, 5), \\ (3, 2, 2, 4, 4, 5, 5, 6, 5, 6, 6, 6, 6, 6, 6, 6, 5).$$