1. Calculate the sum of the side lengths of the 12-gon:

The side lengths are $1, 2, 3, \ldots, 12$. The sum of these side lengths is given by the sum of the first 12 natural numbers:
$$S = \sum_{i=1}^{12} i = \frac{12 \cdot (12 + 1)}{2} = \frac{12 \cdot 13}{2} = 78$$

2. Calculate the sum of all possible triplets of consecutive sides:

There are 12 possible triplets of consecutive sides in a 12-gon. Each triplet can be represented as $(a_i, a_{i+1}, a_{i+2})$ where the indices are taken modulo 12. The sum of the lengths of these triplets is:
$$T = \sum_{i=1}^{12} (a_i + a_{i+1} + a_{i+2})$$
Since each side length appears in exactly three different triplets, the total sum of all triplets is:
$$T = 3 \cdot S = 3 \cdot 78 = 234$$

3. Calculate the average sum of a triplet:

There are 12 triplets, so the average sum of a triplet is:
$$\text{Average sum} = \frac{T}{12} = \frac{234}{12} = 19.5$$

4. **Analyze the possibility of all triplets having a sum $\leq 20$:**

If all triplets had a sum $\leq 20$, then the total sum of all triplets would be at most:
$$12 \cdot 20 = 240$$
However, we have already calculated that the total sum of all triplets is 234. This is less than 240, but we need to consider the distribution of the sums.

5. Consider the distribution of the sums:

Since the average sum is 19.5, if all triplets had sums $\leq 20$, the sums would need to be very close to 20 to achieve the total sum of 234. However, the sums of the triplets are distinct because the side lengths are distinct and arranged in increasing order. Therefore, it is impossible to have a pattern where all triplets are $\leq 20$ without some triplets exceeding 20.

6. **Conclude that at least one triplet must have a sum $> 20$:**

Given the distinct nature of the side lengths and the average sum of 19.5, it is inevitable that at least one triplet must have a sum greater than 20 to balance the sums around the average.

$\blacksquare$