For positive integers $a, n$, define $F_{n}(a)=q+r$, where $q, r$ are non-negative integers, $a=q n+r$, and 0 $\leqslant r<n$. Find the largest positive integer $A$ such that there exist positive integers $n_{1}, n_{2}, n_{3}, n_{4}, n_{5}, n_{6}$, for any positive integer $a \leqslant A$, we have
$$
F_{n_{6}}\left(F_{n_{5}}\left(F_{n_{4}}\left(F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)\right)\right)\right)=1 \text {, }
$$

Prove your conclusion.
To find $A$, we first prove a lemma.

Lemma: If there exists a positive integer $n$, such that for all positive integers $a \leqslant B$ we have $F_{n}(a) \leqslant 2 k$ (where $k$ is a given positive integer), then the largest positive integer $13=k^{2}+3 k$.

Proof: Since $a=q n+r, F_{n}(a)=q+r \leqslant 2 k$, we first do not restrict the range of $r$, and only consider the non-negative integer decompositions of $q+r=i(i=1,2, \cdots, 2 k)$ and their corresponding $a$ values, obtaining Table 1.
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$q+r$ & \multicolumn{5}{|c|}{ Values of natural number $a$ } \\
\hline $2 k$ & $2 k$ & $n+(2 k-1)$ & $2 n+(2 k-2)$ & $\cdots$ & $(2 k-1) n+1$ & $2 k n$ \\
\hline $2 k-1$ & $2 k-1$ & $n+(2 k-2)$ & $2 n+(2 k-3)$ & $\cdots$ & $(2 k-1) n$ & \\
\hline $2 k-2$ & $2 k-2$ & $n+(2 k-3)$ & & $\cdots$ & & \\
\hline $2 k-3$ & $2 k-3$ & & & & & \\
\hline & & & & & & \\
\hline & & & $2 n$ & & & \\
\hline 1 & 1 & $n$ & & & & \\
\hline
\end{tabular}

It is easy to see that the values of $a$ in Table 1 have three important properties:
(1) The numbers in each column increase continuously from bottom to top;
(2) The $i$-th column consists of numbers of the form $(i+1) n+r$;
(3) From the second column onwards, the lowest number in each column is a multiple of $n$.

We call the numbers $q n+r$ in each column of Table 1 that make $r \geqslant n$ "pseudo" numbers. It is easy to see that after removing all the "pseudo" numbers from Table 1, the corresponding $a$ values in the new table are all different. Therefore, for a fixed $n$, the necessary condition for the $a$ values in the new table to cover all natural numbers from 1 to $B(n)$ is that the top number in the $(i+1)$-th column must be $n$ less than 2 (this is because the last number in the $i \div 2$-th column is $(i+1) n$, so there is no number of the form $i n+(n-i)$ in the table), at this time we should have
$$
(i+1)+(n-2)=2 k,
$$

i.e., the maximum value of $a$ is
$$
\begin{array}{l}
B(n)=(2 k+2-n) n+(n-2) \\
=-\left(n-\frac{2 k+3}{2}\right)^{2}+k^{2}+3 k+\frac{1}{4} \\
\leqslant k^{2}+3 k .
\end{array}
$$

Equality holds if and only if $n=k+1$ or $n=k+2$. At this time, Table 2 is given, where $n=k+1$ or $n=k+2$.
Table 2
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$q+r$ & \multicolumn{2}{|c|}{\begin{tabular}{l}
$(2 k+1-n) n+$ \\
$(n-1)$
\end{tabular}} & \begin{tabular}{c}
$(2 k+2-n) n+$ \\
$(n-2)$
\end{tabular} \\
\hline $2 k$ & & & & \begin{tabular}{c}
$(2 k+1-n) n+$ \\
$(n-2)$
\end{tabular} & $\vdots$ \\
\hline $2 k-1$ & & & & & \\
\hline$\vdots$ & & $n+(n-1)$ & $2 n+(n-1)$ & & \\
\hline$n-1$ & $n-1$ & $\vdots$ & & & \\
\hline$\vdots$ & $\vdots$ & $\vdots$ & $2 n+1$ & & & \\
\hline 2 & 2 & $n+1$ & $2 n$ & & & \\
\hline 1 & 1 & $n$ & & & & \\
\hline
\end{tabular}

From Table 2, we can see that $a$ exactly takes all natural numbers from 1 to $k^{2}+3 k$. Thus, the lemma is proved.

Next, we prove that if for all $a \leqslant A$, there exists $n$ such that $F_{n}(a)=1$, then $A=2$.

In fact, $a=q n+r, q+r=1$, so $q=0, r=1$ or $q=1, r=0$, corresponding to $a=1$ or $n$. Clearly, if $n \geqslant 3$, then $A=1$;
if $n=2$, then $A=2$;
if $n=1$, then $A=1$.
Therefore, $A=2$.
Using the above conclusion and repeatedly applying the lemma, we have
$F_{n_{5}}\left(F_{n_{4}}\left(F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)\right)\right)$
$\leqslant 2=2 \times 1$,
$\therefore F_{n_{4}}\left(F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)\right)$
$\leqslant 1^{2}+3 \times 1=4=2 \times 2$,
$F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)$
$\leqslant 2^{2}+3 \times 2=10=2 \times 5$,
$F_{n_{2}}\left(F_{n_{1}}(a)\right) \leqslant 5^{2}+3 \times 5=40=2 \times 20$,
$F_{n_{1}}(a) \leqslant 20^{2}+3 \times 20=460=2 \times 230$.
Therefore, $a \leqslant 230^{2}+3 \times 230=53590$.
Hence, the maximum value of $A$ is 53590, and there exist
$n_{1}=231, n_{2}=21, n_{3}=6, n_{4}=3$,
$n_{5}=2, n_{6}=2$

or $n_{1}=232, n_{2}=22, n_{3}=7, n_{4}=4$,
$n_{5}=3, n_{6}=2$,

such that for any $a \leqslant A$ we have
$F_{n_{6}}\left(F_{n_{5}}\left(F_{n_{4}}\left(F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)\right)\right)\right)=1$.