This problem (prove that the number of Sylow subgroups of a finite group $G$ is bounded by $\frac{2}{3}G$) posted on MSE proved rather difficult to solve. The OP has been silent about where the problem came from, even though he/she has been asked. Has anyone seen this result before? If so, where?

1$\begingroup$ This seems to have appeared as Problem 11856 in American Mathematical Monthly proposed by K. Kearnes (USA). $\endgroup$ – r9m Jul 31 '15 at 21:29

1$\begingroup$ I had a similar experience with AMM11832 that was asked on M.SE on Feb while the problem appeared in May Issue this year. $\endgroup$ – r9m Aug 1 '15 at 15:01

$\begingroup$ @r9m That's why I deleted my comment, I figured it might have been something like that. $\endgroup$ – Matt Samuel Aug 1 '15 at 15:07
Since this question has resurfaced (after 3 years) let me say something about the origin of this Monthly problem.
A few years back a colleague gave me the task of making
up the algebra preliminary exam for our firstyear
graduate students. Among the group theory problems that made
my initial list were:
(1) Show that there is no finite group with more Sylow subgroups than elements.
(This later evolved to: Let $G$ be a finite group. Show that the number of nontrivial Sylow subgroups of $G$ is at most $\frac{2}{3}G$.)
(2) Call a positive integer $c$ curious if there is a nontrivial finite group $G$ such that, for every prime $p$ dividing $G$, the number of Sylow $p$subgroups is exactly $c$. Find all curious integers.
(3) You are told that, for certain $n$ and $p$, $S_n$ has exactly $n$ Sylow $p$subgroups. What are the possible pairs $(n, p)$?
(4) Prove or disprove: There is no group $G$ of squarefree order such that, for every prime $p$ dividing $G$, the number of Sylow $p$subgroups is $p + 1$.
(5) Let $G$ be a finite group and let $S\subseteq G$ be a subset. Show that if $S$ has nonempty intersection with each conjugacy class of $G$, then the subgroup generated by $S$ is $G$.
(6) Show that if the conjugacy classes of $G$ have size at most $4$, then $G$ is solvable.
I ultimately used Problem (6) on the exam. My notes to myself say about (6): Here $4$ can be replaced by any number up to $19$, but the statement becomes false for $20$  $A_5$.
I decided Problem (1) was too hard for firstyear grad students, so I sent it to the problems section of the Monthly. My solution to Problem (1) was similar to the one by Richard Stong, which was published by the Monthly. In particular, I also used Burnside's normal complement theorem and the estimate $\sum_{p}\frac{1}{p^{2}} < \frac{1}{2}$.
Once the problem was typeset by the Monthly, I got a chance to proofread it. I sent my final OK to the Monthly on May 10, 2015. Strangely, the problem appeared on the Art of Problem Solving website the next day, worded in exactly the same way. The day after that it appeared at MSE, worded exactly the same way. It did not appear in print in the Monthly for another 3 months.
One last comment about this problem. When formulating the problem
I asked myself: is it possible to show that there is an injective
function $f:\textrm{Syl}(G)\to G$ such that $f(P)\in P$ for all $P$?
I don't know the answer to this version of the question.
The problem seems to have appeared as Problem 11856 in American Mathematical Monthly in the July 2015 Issue, proposed by K. Kearnes (USA).
This is not particularly helpful, but in the paper by Pyber, Asymptotic results for finite simple groups on page 309 of this conference proceedings, the author states on page 320:
J.P. Zhang proved that the total number of Sylow subgroups of $G$ is at most $G1$ (personal communication).
Strictly speaking the number of Sylow subgroups of $S_3$ is $5$ rather than $4$, because the trivial subgroup is a Sylow $p$subgroup for all primes other than $2$ or $3$, so this bound is in that sense the best possible. But of course it is less good than the $2/3$ bound, and $S_3$ is the only example where this is achieved.

3$\begingroup$ I included a simple proof of $N<G$ (actually $N<(86/100)G)$ as an answer to the MathSE post (for $N=N(G)$ the number of nontrivial Sylow subgroups). It would be interesting to know what is $\limsup_{G\to\infty} N(G)/G$ is. It's in $[1/2,2/3]$ (the lower bound coming from dihedral groups of order $2n$ for $n$ odd). $\endgroup$ – YCor Jun 13 '15 at 10:58

1$\begingroup$ Shouldn't the number of Sylow groups for S_2 be 2 then, strictly speaking? Not to mention 1 for S_1 ? $\endgroup$ – The Masked Avenger Jun 13 '15 at 19:10

$\begingroup$ @TheMaskedAvenger The bound in the linked question is for nontrivial Sylow subgroups. $\endgroup$ – Matt Samuel Jun 13 '15 at 19:29

1$\begingroup$ @MattSamuel , indeed. Which bound was J. Zhang (or is Derek Holt) writing about? $\endgroup$ – The Masked Avenger Jun 13 '15 at 19:32

1$\begingroup$ @TheMaskedAvenger If you allow trivial Sylow subgroups then the total number goes up by one, so the difference is not significant! I have no idea which bound J. Zhang was writing about, or whether he published his proof at all. $\endgroup$ – Derek Holt Jun 14 '15 at 11:28
I consider that it is not at all inappropriate to inform you that a solution to this problem was showcased in the May 2017 issue of the Monthly.
If I understand correctly, the editors of the problems and solutions section of the journal received only two solutions to this problem: the one by the proposer and one by Richard Stong. They chose to showcase Stong's solution: it is a wee bit more succint than the "new, more elementary one" given by Matt Samuel in this MSE thread and it goes as follows:
Let $G$ be a finite group of order greater than $1$. For a prime number $p$ that divides $G$, let $n_{p}(G)$ denote the number of Sylow $p$subgroups of $G$. It is wellknown that if $P$ is a Sylow $p$subgroup of $G$, then $$n_{p}(G) =[G:N_{G}(P)];$$ thus, given that $P \subseteq N_{G}(P)$, it follows that if $p^{m} \parallel G$, then $$n_{p}(G) \leq \frac{G}{p^{m}}.$$
Now then, let $A$ denote the set of all those prime numbers $p$ such that $p \parallel G$. Clearly enough, $A$ consists of those prime divisors $p$ of $G$ such that every Sylow $p$subgroup is cyclic of order $p$. Hence, for every $p\in A$, $G$ has exactly $(p1)n_{p}(G)$ elements of order $p$ and what is more $$\sum_{p \in A} (p1)n_{p}(G) < G.$$
There are two main possibilities to consider:
$\bullet \,$ $A \neq \emptyset$ and $n_{p}(G) = \frac{G}{p}$ for some $p \in A$. We have in this case that any (fixed) Sylow $p$ subgroup $P$ of $G$ is selfnormalizing; since $P$ is a prime number, it follows that $P \subseteq Z(N_{G}(P))$. Then, Burnside's transfer theorem implies the existence of a normal subgroup $H$ of index $p$. Hence, when $q$ is a prime different from $p$, every Sylow $q$subgroup of $G$ is a Sylow $q$subgroup of $H$. If $p \geq 3$, we infer that the number of Sylow subgroups of $G$ is at most (beware of the implicit inductive step)
$$n_{p}(G) + \frac{2}{3}H = \frac{G}{p} + \frac{2G}{3p} = \frac{5G}{3p} < \frac{2}{3}G.$$
If $p=2$, a classical exercise (cf. exercise 15 on page 71 of the second ed. of Yitz Herstein's Topics in algebra) implies that $H$ is an abelian subgroup of $G$. The number of Sylow subgroups of $H$ is equal to the number of prime divisors of $H$; since $H$ is an odd natural number, it follows that the number of different prime divisors of $H$ is at most $H/3$. Thus, the number of Sylow subgroups of $G$ is at most
$$n_{2}(G) +\frac{H}{3}= \frac{G}{2} + \frac{H}{3} = \frac{G}{2} + \frac{G}{6} = \frac{2}{3}G.$$
$\bullet \,$ $A = \emptyset$ or $n_{p}(G) < \frac{G}{p}$ for every $p \in A$. In any of these cases, we have that $$n_{p}(G) \leq \frac{G}{2p}.$$ The inequality $\left(1\frac{2}{p}\right) \leq \frac{p1}{6}$, which is valid for every $p \in \mathbb{N}$, implies in turn that $$n_{p}(G)  \frac{G}{p^{2}}\leq \frac{p1}{6}n_{p}(G)$$ for each $p \in A$. Therefore, the number of Sylow subgroups of $G$ is at most \begin{eqnarray*} \sum_{p \, \mid \, G} n_{p}(G) &=& \sum_{p\in A}n_{p}(G) + \sum_{p \, \mid \, G, \,p \not \in A} n_{p}(G)\\ &\leq & \sum_{p\in A}n_{p}(G) + \sum_{p \, \mid \, G, \, p \not \in A} \frac{G}{p^{2}}\\ &\leq& \sum_{\, p\in A}\left(n_{p}(G)\frac{G}{p^{2}}\right) + \sum_{p}\frac{G}{p^{2}}\\ &\leq& \sum_{\, p\in A}\frac{p1}{6}n_{p}(G) + \sum_{p}\frac{G}{p^{2}}\\ &<& \frac{G}{6} + \sum_{p}\frac{G}{p^{2}}\\ &=& G \left(\frac{1}{6}+\sum_{p}\frac{1}{p^{2}}\right)\\ &<& \frac{2}{3}G \end{eqnarray*} where, in the passage from the next to last to last line, we have resorted to the fact that $$\sum_{p}\frac{1}{p^{2}} \approx 0.452247...$$