Sum of divisors below threshold Let $\sigma(n)$ denote the sum of divisors of $n$, that is,
$$
\sigma(n) = \sum_{d | n} d.
$$
It is known that $\sigma$ can have values as large as order $n \log \log n$. However, obviously the sum is reduced when we restrict it to divisors which are smaller than some threshold $D$.
Question 1: How large can $D$ be (as a function of $n$), such that
$$
\sum_{\substack{1 \leq d \leq D,\\ ~d | n}} d \leq n,
$$
for all (sufficiently large) $n$? 
One admissible value for $D$ is roughly $n / \exp(c \log n / \log \log n)$, since the number of divisors of $n$ is bounded by $\exp(c \log n / \log \log n)$, and the sum above is obviously dominated by $D$ times the number of divisors of $n$. However, my feeling is that the "right" order for $D$ should rather be around $n/\log n$ or even closer to $n$.
Question 2: Can we further increase the bound for $D$ from Question 1 if we only consider those $n$ for which $\varphi(n) > \varepsilon n$, for some fixed constant $\varepsilon$?
 A: Put 
$$
B= C \frac{\log (10\sigma(n)/n)}{\log \log (10 \sigma(n)/n)}
$$ 
for a suitably large positive constant $C$.   Then I claim that the desired inequality holds with 
$$ 
D = \frac{n}{(\log n)^B},
$$ 
for all large $n$.   Since $\sigma(n)/n \ll \log \log n$, it follows that one may always take 
$$ 
D = n \exp\Big( -C \frac{\log_2 n \log_3 n}{\log_4 n}\Big), \tag{1}
$$ 
with $\log_j$ denoting the $j$-th iterated logarithm, and in fact this is in general the best possible.  If $\sigma(n)/n$ is bounded by $1/\epsilon$ (as in Question 2) obviously one can obtain a stronger result, as $B =C \log(1/\epsilon)/\log \log (1/\epsilon)$ is now permissible.
Now for the proof.  Put 
$$
A = \frac 12 \log \log (10\sigma(n)/n), \text{ and  } \alpha= \frac{A}{\log \log n}.
$$ 
Note that (writing $d=n/k$) 
$$ 
\sum_{\substack{ d|n \\ d\le D}} d =n \sum_{\substack{k|n \\ k > (\log n)^B}} \frac{1}{k} \le n (\log n)^{-B\alpha} \sum_{k|n} \frac{1}{k^{1-\alpha}} = ne^{-AB} \sum_{k|n} \frac{1}{k^{1-\alpha}}. \tag{2}  
$$
For large $n$ we have 
$$ 
\sum_{k|n} \frac{1}{k^{1-\alpha}} \ll \prod_{p|n} \Big(1 + \frac{1}{p^{1-\alpha}}\Big) \ll \exp\Big( \sum_{p|n} \frac{1}{p^{1-\alpha}}\Big). 
$$ 
To estimate this, divide the primes $p|n$ into three ranges: $p\le (\log n)^{1/A}$, $(\log n)^{1/A} \le p \le (\log n)$ and $p> \log n$.  The contribution of the first range is 
$$ 
\sum_{\substack{ p|n \\ p\le (\log n)^{1/A}} } \frac{e}{p} \ll \log (\sigma(n)/n). 
$$ 
The second range gives 
$$ 
\le \sum_{(\log n)^{1/A} \le p\le (\log n)} \frac{1}{p^{1-\alpha}} \le e^A \sum_{(\log n)^{1/A} \le p\le (\log n)} \frac 1p \ll e^A \log A.
$$
The final range gives (since the number of distinct primes dividing $n$ is $\ll \log n/\log \log n$) 
$$ 
\ll \sum_{\substack{p|n \\ p>\log n}} \frac{1}{p^{1-\alpha}} \le \frac{1}{(\log n)^{1-\alpha}} \sum_{p|n} 1 \ll \frac{e^A}{\log \log n}.
$$ 
Combining all these bounds, and using it in (2) we find 
$$ 
\sum_{\substack{d|n \\ d\le D}} d \ll n e^{-AB} \exp\Big( O(\log (10\sigma(n)/n))\Big) \le n,  
$$
upon taking $C$ suitably large.  This completes the proof that the claimed choice for $D$ works.  
Let me now quickly say why the general form (1) is optimal.  Take $n$ to be the lcm of all the integers up to some point (which is roughly $\log n$ by the prime number theorem).  Then, roughly speaking, 
$$
\sum_{\substack{k |n \\ k >(\log n)^u}} \frac{1}{k} \gg \sum_{\substack{ (\log n)^{u+1} \ge k \ge (\log n)^u \\ p|k \implies p\le (\log n)}} \frac 1k \gg \rho(u+1) \log \log n, 
$$
where $\rho$ denotes the Dickman function.  That is, the divisors of $n$ basically correspond to $\log n$ smooth numbers, and I have invoked the asymptotic for smooth numbers here.  Since $\rho(u)$ behaves like $u^{-u}$ we see that $u$ has to be about as large as $\log_3 n/\log_4 n$ in order for $\rho(u+1) \log \log n$ to become negligible.  This shows that the range in (1) cannot be improved (apart from the constant $C$). 
