A machine accepts coins of $k{}$ values $1 = a_1 <\cdots < a_k$ and sells $k{}$ different drinks with prices $0<b_1 < \cdots < b_k$. It is known that if we start inserting coins into the machine in an arbitrary way, sooner or later the total value of the coins will be equal to the price of a drink. For which sets of numbers $(a_1,\ldots,a_k;b_1,\ldots,b_k)$ does this property hold?
To solve this problem, we need to determine the sets of coin values \((a_1, a_2, \ldots, a_k)\) and drink prices \((b_1, b_2, \ldots, b_k)\) such that inserting coins in any arbitrary order will eventually match the price of one of the drinks.

1. **Initial Setup and Assumptions:**
   - We are given \(k\) coin values \(1 = a_1 < a_2 < \cdots < a_k\).
   - We are given \(k\) drink prices \(0 < b_1 < b_2 < \cdots < b_k\).
   - The problem states that no matter how we insert the coins, the total value will eventually match one of the drink prices.

2. **Greedy Algorithm and Total Value \(A\):**
   - Suppose we start inserting coins arbitrarily, and let \(A\) be the total value of the coins inserted at some point.
   - According to the problem, there must be a time when the total value \(A\) matches one of the drink prices \(b_j\).

3. **Analyzing the Coin and Drink Sequences:**
   - Since \((a_x)_{x=1}^k\) and \((b_y)_{y=1}^k\) are increasing sequences, we must have \(A + a_i = b_i\) for some \(i\).
   - If \(A = 0\), then \(a_i = b_i\) for all \(i\), implying \((a_1, a_2, \ldots, a_k) = (b_1, b_2, \ldots, b_k)\).

4. **Inductive Step:**
   - Assume \(A \ge 1\). We need to show that the sequences \((a_1, a_2, \ldots, a_k)\) and \((b_1, b_2, \ldots, b_k)\) must follow a specific pattern.
   - By inserting the coin \(a_1 = 1\) \(A-1\) times, we get a total value of \(A-1\).
   - Insert the coin \(a_2\) and then \(a_k\). We know \(A-1 + a_2 < b_2 = A + a_2\).
   - If \(a_2 \ge 3\), then \(A-1 + a_2 > b_1 = A + 1\). Hence, \(A-1 + a_2\) is not equal to the price of a drink.
   - After inserting the coin \(a_k\), the new total value is \(A-1 + a_2 + a_k > A + a_k = b_k\).
   - Thus, we can keep adding the coin \(a_1\) infinitely, and the total value will never match the price of a drink. Therefore, \(a_2 = 2\).

5. **Induction Hypothesis:**
   - Assume \(a_i = i\) for all \(i \leq t-1\) where \(k \ge t \ge 3\).
   - We need to prove \(a_t = t\).
   - By inserting the coin \(a_1 = 1\) \(A-1\) times, we get a total value of \(A-1\).
   - Insert the coin \(a_t\) and then \(a_k\). We know \(A-1 + a_t < b_t = A + a_t\).
   - If \(a_t \ge t+1\), then \(A-1 + a_t > b_{t-1} = A + a_{t-1} = A + t-1\). Thus, \(A-1 + a_t\) is not equal to the price of a drink.
   - After inserting the coin \(a_k\), the new total value is \(A-1 + a_t + a_k > A + a_k = b_k\).
   - Thus, we can keep adding the coin \(a_1\) infinitely, and the total value will never match the price of a drink. Hence, \(a_t = t\).

6. **Conclusion:**
   - By induction, we have shown that \(a_i = i\) for all \(i \leq k\).
   - Therefore, the sequences must be \((a_1, a_2, \ldots, a_k) = (1, 2, \ldots, k)\) and \((b_1, b_2, \ldots, b_k) = (n+1, n+2, \ldots, n+k)\) for some integer \(n\), or \((a_1, a_2, \ldots, a_k) = (b_1, b_2, \ldots, b_k)\).

The final answer is \( \boxed{ (a_1, a_2, \ldots, a_k) = (1, 2, \ldots, k)\; ;\; (b_1, b_2, \ldots, b_k) = (n+1, n+2, \ldots, n+k)\text{ or }(a_1, a_2, \ldots, a_k) = (b_1, b_2, \ldots, b_k) } \).