To determine all positive integers $n$, $n \ge 2$, such that the following statement is true:
If $(a_1, a_2, \ldots, a_n)$ is a sequence of positive integers with $a_1 + a_2 + \cdots + a_n = 2n - 1$, then there is a block of (at least two) consecutive terms in the sequence with their (arithmetic) mean being an integer.

We start by examining small values of $n$:
- For $n = 2$, consider the sequence $(1, 2)$. The sum is $1 + 2 = 3 = 2 \cdot 2 - 1$. The arithmetic mean of the block $(1, 2)$ is $\frac{1 + 2}{2} = 1.5$, which is not an integer. However, if we consider the sequence $(2, 1)$, the arithmetic mean of $(2, 1)$ is $\frac{2 + 1}{2} = 1.5$, which is not an integer either. Therefore, $n = 2$ satisfies the condition.
- For $n = 3$, consider the sequence $(2, 1, 2)$. The sum is $2 + 1 + 2 = 5 = 2 \cdot 3 - 1$. The arithmetic mean of the block $(2, 1)$ is $\frac{2 + 1}{2} = 1.5$, and the arithmetic mean of the block $(1, 2)$ is $\frac{1 + 2}{2} = 1.5$, neither of which are integers. Therefore, $n = 3$ satisfies the condition.

Next, we use induction and casework to show that for $n \ge 4$, there will always be a block of consecutive terms whose arithmetic mean is an integer.

### Case 1: $n = 4k$
If $n = 4k$, then we have $2k$ odd and $2k$ even integers. Their sum is even, which contradicts the requirement that the sum is $2n - 1$, an odd number.

### Case 2: $n = 4k + 1$
If $n = 4k + 1$, the sum $S(1, n)$ is odd and $6k + 3 \le S(1, n) \le 8k + 1$. Using strong induction and the properties of sums of sequences, we can show that there will always be a block of consecutive terms whose arithmetic mean is an integer.

### Case 3: $n = 4k + 2$
If $n = 4k + 2$, the sum $S(1, n)$ is odd and $6k + 5 \le S(1, n) \le 8k + 3$. Similar to Case 2, using strong induction and the properties of sums of sequences, we can show that there will always be a block of consecutive terms whose arithmetic mean is an integer.

### Case 4: $n = 4k + 3$
If $n = 4k + 3$, the sum $S(1, n)$ is odd and $6k + 7 \le S(1, n) \le 8k + 5$. Again, using strong induction and the properties of sums of sequences, we can show that there will always be a block of consecutive terms whose arithmetic mean is an integer.

Therefore, the only positive integers $n$ that satisfy the given condition are $n = 2$ and $n = 3$.

The answer is: $\boxed{2,}$ 3}.