Suppose that he paid the sum of $S$ UAH with $k$ banknotes, among which there are $a$ of 3 UAH, $b$ of 25 UAH, and $c$ of 80 UAH. Then
$$S = 3a + 25b + 80c \equiv 3a + 3b + 3c = 3k \pmod{11}.$$
If $n+1$ is the sum that he paid in the first day, then in the $i$-th day he paid a sum of $n+i$ with precisely $k_i$ banknotes. From the above formula it follows that $3(k_1 + k_2 + \cdots + k_{2024}) \equiv (n+1) + (n+2) + \cdots + (n+2024) = 2024n + 1012 \cdot 2025 \pmod{11}$.
As $1012 \equiv 11$, the right side is divisible by 11, so the left side, equal to the total number of pair banknotes, also is divisible by 11. But the number 1000000 isn't divisible by 11, so Victor couldn't have used exactly a million banknotes.