Best upper bound on the number of divisors of $n$ that are larger than $N$. I am looking for the best upper bound on $$\sum_{\substack{d | n\\ d \geq N}} 1.$$ 
I know that
$$
d(n) = \sum_{\substack{d | n}} 1 \leq e^{O(\frac{\log n}{\log \log n})}.
$$
For my application, I need something like 
$$\sum_{\substack{d | n\\ d \geq N}} 1 \leq \frac{o(n^{\epsilon})}{\log N} \quad \forall \epsilon > 0.
$$ 
A reference where the bound can be found or a simple proof would be appreciated.
Thanks.
EDIT: Johan Andersson has pointed out that the third display follows from the second. (Thanks.) I am still interested to learn what the best known bound is.
 A: This is not an answer to what the best known upper bound is, but rather a comment that the (known) average distribution of divisors indicates you might not expect to do any better than the bounds on the divisor function itself.  Tennenbaum's "Introduction to Analytic and Probabilistic Number Theory", $\S$ 6.2 p. 207 says

For each integer $n$, let us define a
  random variable $D_n$ taking the
  values $\log d/\log n$, as $d$ runs
  through the the set of the $\tau(n)$
  divisors of $n$, with uniform
  probability $1/\tau(n)$.  The
  distribution function $F_n$ of $D_n$
  is then defined by $$
> F_n(u):=\text{Prob}(D_n\le
> u)=\frac{1}{\tau(n)}\sum_{d|n,d\le
> n^u}1\quad (0\le u\le 1). $$ It is
  clear that the sequence
  $\{F_n\}_{n=1}^\infty$ does not
  converge pointwise on $[0,1]$. 
  However, we shall see the sequence of
  Cesaro means $$
> G_N(u):=\frac{1}{N}\sum_{n\le N}F_n(u)
> $$ is uniformly convergent on $[0,1]$.
  Remarkably, the limit is the
  distribution function of a probability
  law well known to specialists: the
  arcsine law, with density
  $1/(\pi\sqrt{u(1-u)})$.  Large and
  small values have high probability: if
  $D$ is a random variable with this
  distribution law, we have $$
> \text{Prob}(D<0.01\text{ or
> }D>0.99)\approx0.128 $$ This indicates
  that, on average, an integer has many
  small (and correspondingly many large)
  divisors.

Update:  One thing we learn is that the relevant parameter is not $N$, but $u:=\log_n(N)$.
