I can solve the problem if by exceeds we mean $\geq$ rather than $>$. Perhaps someone can extend my ideas: Firstly, if $b-a \geq 1/6$ then we are done, assume therefore that $b-a< 1/6$. It follows that in the interval $[a,b]$ we cannot have BOTH rational numbers of the form $n/3$ and $n/2$ since otherwise our interval's length will be greater than or equal to $1/6$. So if we have no number of the form $n/3$, then the interval $[3a,3b]$ contains no integers. On the other hand, if we have no number of the form $n/2$ then we can consider the interval $[2a,2b]$, it has no integer in it. We can then continue this process until our interval is of length greater than or equal to $1/6$. edit: actually, I am right, we need exceed to mean $\geq$ rather than $>$. Because consider the interval $[1/3, 1/2]$It has length $1/6$ therefore $N=1$ is not enough. So can $N=2$? no because then 1 is in the interval can $N=3$? no because then 1 is also in the interval. can $N=4$? no because then 2 is in the interval. can $N=5$? no because then 2 is in the interval since $1/3 < 2/5 < 1/2$. can $N=6$? no because then $2$ and $3$ are in our interval. $N>6$ is not possible since then our interval length is greater than 1, and hence contains an integer.

$N$ needn't be an integer; choosing, for instance, $N = \frac{3}{2}$ for $[\frac{1}{3}, \frac{1}{2}]$ satisfies the condition.

Ok then if that's the case then you can easily add those sorts of cases in (just multiply those intervals of length exactly $1/6$ by some number a little bit bigger than 1 and you are done). I thought $N$ is an integer because otherwise the problem may be too simple. I showed that $1/6$ is best possible for integer $N$(with the extra condition that we mean $\geq$ by exceeds) if we let $N$ be any positive real number as you say would it also be best possible?

I think that this problem might be meaningless. If $N$ is not an integer,we let $N=\frac{1}{6\sqrt{a^2+b^2}}$,obviously we have $\frac{b}{6\sqrt{a^2+b^2}}-\frac{a}{6\sqrt{a^2+b^2}}=\frac{1}{6}(\sin{\alpha}-\cos{\alpha})<\frac{1}{6}$. If $N$ is an integer,when we choose $(0.1,0.9)$ ,obviously whatever $N\geq2$ is,the length of the interval is greater than $1$ and certainly contain integers.Thus it must be wrong if $N$ is not an integer.

If any real $N$ is allowed, the critical value is $\dfrac {1} {2}$, not $\dfrac {1} {6}$. Moreover, since this threshold is not reachable, we will work with open intervals. We may assume $b> a > 0$ (otherwise multiply everything by $-1$). The example of the interval $\left(\dfrac {1} {2} - \varepsilon, 1\right)$ shows we cannot enlarge intervals of length more than $\dfrac {1} {2}$. For any interval $(a,b)$ of length less than $\dfrac {1} {2}$, let us multiply by the largest $N$ such that $Nb$ becomes some integer $k \geq 1$, while $N(b-a) \leq 1$. Then by multiplying with $N\dfrac {k+1} {k} > N$ we should have $N\dfrac {k+1} {k}(b-a) > 1$, so $N(b-a) > \dfrac {k} {k+1} \geq \dfrac {1} {2}$, so we obtained an interval of length more than $\dfrac {1} {2}$. In order to regain the problem on closed intervals, we can always adjust by some final multiplier close enough to $1$ so we avoid integers at the ends of the interval, while keeping the length of the interval as close as wanted to the just previous value. Therefore I assume Valentin meant $N$ integer ...