Anna and Béla play the following game: Anna thinks of a quadruple of natural numbers $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$. Béla states a quadruple of natural numbers $\left(b_{1}, b_{2}, b_{3}, b_{4}\right)$. In response, Anna tells him the sum $a_{1} b_{1}+a_{2} b_{2}+$ $a_{3} b_{3}+a_{4} b_{4}$. Then Béla can state more quadruples. How many quadruples does Béla need at least to find out Anna's numbers?
Two number quadruples are sufficient to guess Anna's numbers. If Béla chooses the first number quadruple as $(1,1,1,1)$ and the answer to this is a $k$-digit number, then it follows that Anna thought of a number quadruple consisting of at most $k$-digit numbers. Thus, if the second number quadruple is $1,10^{k}, 10^{2k}, 10^{3k}$, then the last $k$, the $k$ digits before the last $k$, etc., of the at most $4k$-digit number Anna says will be the first, second, etc., numbers of the guessed number quadruple. However, a single number quadruple is generally insufficient. If Anna's answer to an arbitrary $b_{1}, b_{2}, b_{3}, b_{4}$ number quadruple is $b_{1}b_{2} + b_{1} + b_{2} + b_{3} + b_{4}$, then it cannot be determined whether Anna thought of $(1 + b_{2}, 1, 1, 1)$ or $(1, 1 + b_{1}, 1, 1)$. Therefore, Béla can always guess Anna's numbers based on two number quadruples, but generally less is not sufficient.

Remarks. 1. Most of the submitters overlooked the fact that Béla can choose his second number quadruple based on the knowledge of the first answer, and that the numbers Anna thinks of must be natural numbers. Thus, many were satisfied with the conclusion based on the principle of "n unknowns in n equations," that Béla can determine Anna's numbers from any four number quadruples he arbitrarily chooses. There are still more gaps in this "solution"! On the one hand, if Béla chooses the same number quadruple or multiples of the same number quadruple, then the answer will also be a multiple of the same number, so the equations Béla gets are multiples of each other, and he cannot determine the number quadruple. On the other hand, it must also be shown that fewer number quadruples are not sufficient. It must be ensured that the number quadruples Béla says are such that none of them can be expressed as the sum of multiples of the others, i.e., the equations are "independent." It must be shown that in this case the unknowns are always determinable, and if there are fewer equations, they are generally not. These problems were partially avoided or solved by those who wrote that if Béla says the coefficients $(1,1,1,1), (2,1,1,1), (1,2,1,1), (1,1,2,1)$, then he can determine Anna's numbers from four questions (of course, this gives the solution to a modified problem).

2. Unfortunately, most of those who realized that two questions are sufficient did not show that this number is also necessary.
3. It follows from the solution that any number $n$-tuple can also be guessed with two questions. However, if negative or fractional numbers are also allowed, then our method does not work.