Problem 1972

Answer: for all $n \ge 3$.

Assume $n \ge 3$. Let us introduce such denotations:
$$a_1 + a_2 + \dots + a_{n-1} = a, \quad a_1 a_2 \dots a_{n-1} = A, \quad b_1 + b_2 + \dots + b_{n-1} = b, \quad b_1 b_2 \dots b_{n-1} = B. \quad (1)$$
Then we can write that $a + a_n = b + b_n$, $A a_n = B b_n$, and hence,
$$\begin{aligned} \frac{A a_n}{B} &= b_n \Rightarrow a + a_n = b + \frac{A a_n}{B} \Rightarrow B a_n - A a_n = b B - a B \Rightarrow \\ a_n &= \frac{B(b-a)}{B-A} \Rightarrow b_n = \frac{A(b-a)}{B-A}. \end{aligned} \quad (2)$$
Thus, having a set of $2n-2$ numbers that satisfy conditions (1), one can always add numbers $a_n$ and $b_n$ determined by formulas (2). Then a set of $2n$ numbers satisfies the condition if and only if the signs of the numbers $(b-a)$ and $(B-A)$ are the same.
As an example of the opposite, consider the numbers $a_1 = 3, a_2 = 5, b_1 = 2, b_2 = 7$. Then $a=8, b=9, A=15, B=14$. Then $a_3 = \frac{B(b-a)}{B-A} = -14, b_3 = -15$. Indeed, the sets of numbers
3; 5; -14 and 2; 7; -15 have the same sums and products, but are constituted by not only positive integers.

The easiest way is to choose numbers such that the first $2n-2$ numbers had products differing by 1, i.e. $|B-A|=1$. Then numbers found by formulas (2) will be integer, and if they satisfy other conditions, they will meet the conditions of the problem. Thus let us find two numbers, each of which has a different decomposition into at least $n-1$ factors, and which are neighbouring positive integers. As the basis let us choose $2^N - 1$ and $2^N$.
Let us choose numbers so that:
$$a_1 = 2^{2^1} - 1 = 3;$$
$$a_2 = 2^{2^1} + 1 = 5, \text{ then } a_1 a_2 = 2^{2^2} - 1;$$
$$a_3 = 2^{2^2} + 1 = 17, \text{ then } a_1 a_2 a_3 = 2^{2^3} - 1; \dots$$
$$a_{n-2} = 2^{2^{n-3}} + 1, \text{ then } a_1 a_2 \dots a_{n-2} = 2^{2^{n-2}} - 1;$$
$$a_{n-1} = 2^{2^{n-2}} + 1, \text{ then } a_1 a_2 \dots a_{n-1} = 2^{2^{n-1}} - 1;$$
$$b_1 = 2^{2^0} = 2;$$
$$b_2 = 2^{2^1} = 4, \text{ then } b_1 b_2 = 2^{2^0 + 2^1} = 2^{2^2 - 1};$$
$$b_3 = 2^{2^2} = 16, \text{ then } b_1 b_2 b_3 = 2^{2^3 - 1}; \dots$$
$$b_{n-2} = 2^{2^{n-3}}, \text{ then } b_1 b_2 \dots b_{n-2} = 2^{2^{n-2}-1};$$
$$b_{n-1} = 2^{2^{n-2}+1}, \text{ then } b_1 b_2 \dots b_{n-1} = 2^{2^{n-1}}.$$
Due to the construction we have that $b_1 b_2 \dots b_{n-1} = B = 2^{2^{n-1}}$, $a_1 a_2 \dots a_{n-1} = A = 2^{2^{n-1}} - 1$. What is left to check other conditions. $B = A+1 > A$, hence it should be true that $b > a$.
$$a = a_1 + a_2 + \dots + a_{n-1} = 2^{2^1} - 1 + 2^{2^1} + 1 + 2^{2^2} + 1 + \dots + 2^{2^{n-3}} + 1 + 2^{2^{n-2}} + 1,$$
$$b = b_1 + b_2 + \dots + b_{n-1} = 2^{2^0} + 2^{2^1} + \dots + 2^{2^{n-3}} + 2^{2^{n-2}+1}.$$
$$b-a = 2^{2^0} + 2^{2^1} + \dots + 2^{2^{n-3}} + 2^{2^{n-2}+1} - 2^{2^1} - 2^{2^1} - 2^{2^2} - \dots - 2^{2^{n-3}} - 2^{2^{n-2}} - (n-3) = \\ = 2^{2^0} - 2^{2^1} + 2^{2^{n-2}+1} - 2^{2^{n-2}} - (n-3) = 2^{2^{n-2}} - (n-1) > 0$$
The last inequality can be proved easily by Mathematical Induction.
It remains to check that the numbers are pairwise different. I.e., that the added numbers $a_n$ and $b_n$ do not equal any of the numbers constructed earlier. Provided that $B-A=1$, we obtain
$$a_n = \frac{B(b-a)}{B-A} = B(b-a) > b_n = \frac{A(b-a)}{B-A} = A(b-a) = (2^{2^{n-1}} - 1)(2^{2^{n-2}} - (n-1)) > \\ b_{n-1} = 2^{2^{n-2}+1} > a_{n-1} = 2^{2^{n-2}} + 1.$$
Let us prove that $2^{2^{n-1}} - 1 > 2^{2^{n-2}+1}$, and even a stronger inequality: $2^{2^{n-1}} \ge 2^{2^{n-2}+2}$. Then we have:
$$2^{2^{n-1}} \ge 2^{2^{n-2}+2} \Leftrightarrow 2^{n-1} \ge 2^{n-2} + 2 \Leftrightarrow 2^{n-2} \ge 2.$$
For $n=1$ it is clear that such numbers do not exist.

For $n=2$, if $a_1 a_2 = b_1 b_2 = q$ and $a_1 + a_2 = b_1 + b_2 = p$, then the numbers $a_1, a_2$ and $b_1, b_2$ are the roots of the square equation $t^2 - p t + q = 0$, and therefore these pairs of numbers are the same up to the order.

Alternative solution.
Suppose we have $2n-2$ pairwise different numbers $a_1, a_2, \dots, a_{n-1}$, $b_1, b_2, \dots, b_{n-1}$, which satisfy the conditions:
$$a_1 + a_2 + \dots + a_{n-1} = a = b_1 + b_2 + \dots + b_{n-1} - 1, \quad a_1 a_2 \dots a_{n-1} = A = b_1 b_2 \dots b_{n-1} - 1. \quad (3)$$
Thus, for the formulas (2) we have $b-a=1$ and $B-A=1$, therefore we can make a set of $2n$ pairwise different numbers $a_1, a_2, \dots, a_{n-1}, a_n, b_1, b_2, \dots, b_{n-1}, b_n$ by adding two numbers obtained by the formulas (2), $a_n = B = A+1$ and $b_n = A$.
What is left is to understand how for every positive integer $n \ge 2$ obtain two sets of $n$ numbers, sums and products of which differ by 1.
Suppose pairwise different numbers $a_1, a_2, \dots, a_{n-1}, b_1, b_2, \dots, b_{n-1}$ satisfy conditions (3). Let us add to them numbers: $a_n = A$ and $b_n = A-1$. Then
$$a_1 + a_2 + \dots + a_{n-1} + a_n = a + A, \quad b_1 + b_2 + \dots + b_{n-1} + b_n = a + 1 + A - 1 = a + A, \\ a_1 a_2 \dots a_{n-1} a_n = A A = A^2, \quad b_1 b_2 \dots b_{n-1} b_n = (A+1)(A-1) = A^2 - 1.$$
Now let us add $a_{n+1} = A^2 - 2$ and $b_{n+1} = A^2 - 1$ to these numbers and obtain:
$$a_1 + a_2 + \dots + a_{n-1} + a_n + a_{n+1} = a + A + A^2 - 2, \\ b_1 + b_2 + \dots + b_{n-1} + b_n + b_{n+1} = a + A + A^2 - 1, \\ a_1 a_2 \dots a_{n-1} a_n a_{n+1} = A^2 \cdot (A^2 - 2) = A^4 - 2A^2, \\ b_1 b_2 \dots b_{n-1} b_n b_{n+1} = (A^2 - 1)(A^2 - 1) = A^4 - 2A^2 + 1.$$
As you can see, we made sets of $2n+2$ pairwise different numbers $a_1, a_2, \dots, a_{n+1}, b_1, b_2, \dots, b_{n+1}$, that satisfy conditions (3), therefore from them one can obtain the desired sets of $2n+4$ numbers. Thus, from the existence of sets satisfying conditions (3) for $n=2$, we obtain that there are desired sets of numbers for any odd $n \ge 3$, and from existence of sets satisfying conditions (3) for $n=3$ it follows that there are sets for any even $n \ge 4$. What is left is to state these numbers.
For $n=2$ we can take $a_1 = 4, a_2 = 5, b_1 = 3, b_2 = 7$,
For $n=3$ we can take $a_1 = 2, a_2 = 3, a_3 = 15, b_1 = 1, b_2 = 7, b_3 = 13$.
Then, e.g., we make sets as follows:
For $n=3$: $a_1 = 4, a_2 = 5, a_3 = 21$ and $b_1 = 3, b_2 = 7, b_3 = 20$.
With the set for $n=2$ satisfying conditions (3), we now make a set for $n=4$ that also satisfies the conditions (3):
$$a_1 = 4, a_2 = 5, a_3 = 20, a_4 = 398 \text{ and } b_1 = 3, b_2 = 7, b_3 = 19, b_4 = 399. \text{ Then we have that} \\ a_1 + a_2 + a_3 + a_4 = 427, b_1 + b_2 + b_3 + b_4 = 428, \text{ and} \\ a_1 a_2 a_3 a_4 = 159200, b_1 b_2 b_3 b_4 = 159201.$$