Why do you think so?
The following is not an induction approach.
Suppose $x=a+y$ where $a\in\mathbb{Z}$ and $0\le y<1$, i.e., $\lfloor x\rfloor=a$. Also let $k=0,\dots,n-1$ be such that
$$
y+k/n<1,\text{ but }y+(k+1)/n\ge1.\qquad(1)
$$
This means that $\lfloor z\rfloor=a$ for $z=x,\dots,x+k/n$ (the first $k+1$ terms of the left-hand side), but $\lfloor z\rfloor=a+1$ for $z=x+(k+1)/n,\dots,x+(n-1)/n$ (the last $n-k-1$ terms of the left-hand side). Thus, the left-hand side is $(k+1)a+(n-k-1)(a+1)= na+n-k-1$. Now derive lower and upper bounds on $ny$ from (1) and use it to find $\lfloor nx\rfloor$.
How? I typed it.
Are you asking why such nonnegative integer k exists? Take k = 0. Then $y+k/n=y<1$ by assumption about y. If already $y+1/n\ge1$, then the k that I am talking about is 0 because $y+k/n<1$, but $y+(k+1)/n\ge1$. If $y+1/n$ is still less than 1, then consider $y+2/n$ and so on. Eventually you will find such k because you start when the number (i.e., y) is < 1, and as you keep adding 1/n to it, eventually the sum will exceed 1. The idea is to denote by k + 1 the number of terms from the original sum that are rounded to $\lfloor x\rfloor$ (the rest n - k - 1 are rounded to $\lfloor x\rfloor+1$).
I proved for $1\leq i\leq n$ that $\lfloor x+{n-i\over n}\rfloor$ is either 0 or 1 and is 1 precisely for $1\leq i\leq\lfloor nx\rfloor$. So
$$\sum_{i=1}^n\lfloor x+{n-i\over n}\rfloor=1+1+\cdots+1+0+0+\cdots+0$$
There are precisely $\lfloor nx\rfloor$ 1's in the above sum.