Solution:

We will prove that the minimum number of measurements equals 2. Denote the weights of a genuine and a counterfeit coin by $x$ and $y$ respectively, and let $a_{i}=1$ if the coins in the $i$-th wagon are counterfeit, and $a_{i}=0$ otherwise.

In the first measurement, take one coin from each wagon. Then $a_{1}+a_{2}+\cdots+a_{n} = \frac{n x - m_{1}}{x-y}$, where $m_{1}$ is the obtained mass. We assume that $m_{1} \neq n x$, since otherwise there are no counterfeit coins.

In the second measurement, for some $q \in \mathbb{N}$, we take $q^{i-1}$ coins from the $i$-th wagon. If the obtained mass is $m_{2}$, we have $a_{1} + q a_{2} + \cdots + q^{n-1} a_{n} = \frac{(1+q+\cdots+q^{n-1}) x - m_{2}}{x-y}$. From here we obtain
$$f\left(a_{1}, a_{2}, \ldots, a_{n}\right) = \frac{a_{1} + q a_{2} + \cdots + q^{n-1} a_{n}}{a_{1} + a_{2} + \cdots + a_{n}} = \frac{(1+q+\cdots+q^{n-1}) x - m_{2}}{n x - m_{1}}$$
We want the value of $f$ to uniquely determine $a_{1}, \ldots, a_{n}$. Thus, it suffices to show that there exists a natural number $q$ such that the function $f:\{0,1\}^{n} \setminus \{(0,0, \ldots, 0)\} \rightarrow \mathbb{R}$ is injective.

For fixed $\mathfrak{a} = (a_{1}, a_{2}, \ldots, a_{n})$ and $\mathfrak{b} = (b_{1}, b_{2}, \ldots, b_{n})$, the equality $f(\mathfrak{a}) = f(\mathfrak{b})$ is equivalent to $P_{\mathfrak{a}, \mathfrak{b}}(q) = (a b_{n} - b a_{n}) q^{n-1} + \cdots + (a b_{2} - b a_{2}) q + (a b_{1} - b a_{1}) = 0$, where $a = a_{1} + a_{2} + \ldots + a_{n} \neq 0 \neq b = b_{1} + b_{2} + \ldots + b_{n}$. Therefore, if the function $f$ is not injective, then $q$ is a root of the polynomial
$$P(q) = \prod_{\mathfrak{a}, \mathfrak{b}} P_{\mathfrak{a}, \mathfrak{b}}(q)$$
Since none of the polynomials $P_{\mathfrak{a}, \mathfrak{b}}$ is identically equal to 0, there are only finitely many numbers $q$ for which $P(q) = 0$, so it is possible to choose $q$ for which $f$ is an injective function.

In this way, in two measurements we can determine $a_{1}, \ldots, a_{n}$, i.e., the wagons with counterfeit coins. Finally, we determine $y$ from the equality $y = x - \frac{n x - m_{1}}{a_{1} + a_{2} + \cdots + a_{n}}$.

On the other hand, one measurement is not sufficient, because by taking $k_{i}$ coins from the $i$-th wagon we obtain the equation $k_{1} a_{1} + \cdots + k_{n} a_{n} = \frac{k x - m}{x-y}$ (where $k = k_{1} + \cdots + k_{n}$) which in general has multiple solutions. For example, two possible solutions for $(a_{1}, a_{2}, \ldots, a_{n}, y)$ are $(1, 0, \ldots, 0, x - \frac{k x - m}{k_{1}})$ and $(0, 0, \ldots, 1, x - \frac{k x - m}{k_{n}})$.