We will prove a more general result about the solutions of equations of the type ⌊x⌋+⌊2x⌋+...+⌊nx⌋=n+1.
First, we look at the case n=3k.
We denote the term on the LHS by f(k,x).
We first see that for any fixed k, f(k,x) is increasing in x.
Also, f(k,1k)=k+k+k+3=3k+3>3k+1 by trivial calculation.
Let's look at f(k,1k−ϵ). For small ϵ, almost all the terms of f(k,1k−ϵ) are the same as the corresponding ones of f(k,1k except for ⌊kx⌋,⌊2kx⌋ and ⌊3kx⌋ who are one less than before. Hence, for sufficiently small ϵ we have f(k,1k−ϵ)=3k<3k+1. Hence, there can't be any real x such that f(k,x)=3k+1 which also solves part b).
In a similar way, one can prove that the equation has solutions when n is not divisible by 3. They are given by 22k+1≤x<33k+1 in the case n=3k+1 and 22k+2≤x<33k+2 in the case n=3k+2. The special case for k=670 (part a)) now follows easily...