Sadly solved after the revenge. We generalize to the following problem: Claim. Let $a,b$ be positive integers. For each natural $m$ and each integer sequence $a\le k_1<k_2<\cdots <k_m$ for which $k_1+\ldots+k_m=b$, Michael wrote down the number $\frac{1}{k_1\cdot k_2\cdots k_m} $ on the board. Then the sum of the numbers on the board is at most $\frac{1}{a}$, with equality iff $a=b$. Proof. The proof will be by strong induction on $b-a$. Let $f(a,b)$ be the sum of the numbers on the board. For the base case, assume $b-a\le 0$. If $b<a$ then $f(a,b)=0<\frac{1}{a}$. If $b=a$ then $f(a,b)=\frac{1}{a}$. Now assume $b-a=k\ge 1$ and the claim holds whenever $b-a<k$. Observe that $$f(a,b)=\frac{1}{a}f(a+1,b-a)+f(a+1,b).$$This is proved by considering cases whether $k_1=a$ or $k_1\ge a+1$ in the equation $k_1+\ldots +k_m=b$. Since $(b-a)-(a+1)<b-a$ and $b-(a+1)<b-a$, induction hypothesis implies that $$f(a,b)=\frac{1}{a}f(a+1,b-a)+f(a+1,b)\le\frac{1}{a(a+1)}+\frac{1}{a+1}=\frac{1}{a}.$$For equality we need that simultanously $a+1=b-a$ and $a+1=b$. Hence $a=0$, which is a contradiction since $a$ is a positive integer. Therefore, equality cannot hold so $f(a,b)<\frac{1}{a}$. This completes the induction. $\square$ The problem follows from the claim by substituting $(a,b)=(1,n)$.

Here is a slick solution by Omri Nisan Solan. Lemma. Let $n$ be a positive integer. Let $k_1<k_2<\cdots <k_m$ be positive integers such that $\sum_{i=1}^m k_i=n$. Then the number of permutations of $\{1,\ldots,n\}$ that are products of $m$ disjoint cycles of lengths $k_1,k_2,\ldots,k_m$ is exactly $\frac{n!}{\prod_{i=1}^m k_i}$. Proof. There are exactly $\binom{n}{k_1,\ldots,k_m}$ ways to partition $\{1,\ldots,n\}$ into $m$ disjoint subsets of sizes $k_1,\ldots,k_m$. For each subset of size $k_i$, there are exactly $(k_i-1)!$ permutations over that subset which are full cycles. Hence the number of permutations of $\{1,\ldots,n\}$ that are products of $m$ disjoint cycles of lengths $k_1,k_2,\ldots,k_m$ equals $$\binom{n}{k_1,\ldots,k_m}\prod_{i=1}^m (k_i-1)!=\frac{n!}{\prod_{i=1}^m k_i}$$as desired. $\square$ From the lemma, it follows that the number of permutations of $\{1,\ldots,n\}$ which are products of disjoint cycles with pairwise distinct lengths is exactly $$\sum_m\sum_{\begin{matrix}0<k_1<\ldots<k_m\\k_1+\ldots+k_m=n\end{matrix}}\frac{n!}{k_1k_2\cdots k_m}.$$However, when $n\ge 2$, the identity permutation is not a product of disjoint cycles with pairwise distinct lengths. Hence the result is strictly less than $n!$. Dividing by $n!$ gives the original problem statement.

Notice that for $n\ge 2$,we have $\sum_{m\ge 1}\sum_{\begin{matrix}0<k_1<\ldots<k_m\\k_1+\ldots+k_m=n\end{matrix}}\frac{1}{k_1k_2\cdots k_m}=[x^{n}]\prod_{k=1}^{+\infty}(1+\frac{x^{k}}{k})$ $<[x^{n}]\prod_{k=1}^{+\infty}\left(\sum_{\ell =1}^{+\infty}\frac{1}{\ell !}\left(\frac{x^{k}}{k}\right)^{\ell}\right)$. Actually,$\prod_{k=1}^{+\infty}\left(\sum_{\ell =0}^{+\infty}\frac{1}{\ell !}\left(\frac{x^{k}}{k}\right)^{\ell}\right)=\prod_{k=1}^{+\infty}\text{e}^{\frac{x^{k}}{k}}=\text{e}^{\sum_{k=1}^{+\infty}\frac{x^{k}}{k}}=\text{e}^{-\ln (1-x)}=\frac{1}{1-x}=1+x+x^{2}+\cdots$. Therefore,for $n\ge 2$,we have $\sum_{m\ge 1}\sum_{\begin{matrix}0<k_1<\ldots<k_m\\k_1+\ldots+k_m=n\end{matrix}}\frac{1}{k_1k_2\cdots k_m}<[x^{n}](1+x+x^{2}+\cdots)=1$.