Let $n$ be a non-negative integer; $M=\{0,1,2, \cdots\}$, then the following table defines $f_n$. Here, $a_{1}$ equals the number of times $k$ appears in $a_{0}, a_{1}, \cdots, a_{\mathrm{n}}$. For example: when $n=3$, $f_{3}$ can be defined as in the table on the right.
Determine for which non-negative integers $n, f_{n}$ does not exist? And explain the reason.

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Because $a_{1} \geqslant 0$, each $a_{1}$ represents the number of times $i$ appears in the corresponding sequence, i.e., $\Omega_{1}$ boxes are filled with $i$, and there are a total of $n+1$ boxes, so
$$
\begin{array}{l}
a_{0}+a_{1}+a_{2}+\cdots+a_{n}=n+1, \\
\left.a_{0} \times 1\right)+a_{1} \times 1+a_{2} \times 2+\cdots+a_{0} \times n \\
=n+1 .
\end{array}
$$

When $n \geqslant 6$, construct the following correspondence table:
From $p+2+1+1=n+1, p+3=n$,
so $p=n-3 \geqslant 3$, and $n \geqslant 6$, hence when $n \geqslant 6$, $f_{\mathrm{n}}$ always exists.
When $n=5$,
$$
a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=6 .
$$

It is easy to determine that $a_{0} \leqslant 5$.
Discussing $a_{0}=1,2,3,4,5$ respectively (the proof is left to the reader) all lead to contradictions, so
$f_{0}$ does not exist.
When $n=4$,

there is Figure 1, so $f_{4}$
exists.
When $n=3$,

there is Figure 2 or Figure 3, so
$f_{s}$ exists.
It is easy to prove that for $n=2$,
$n=1, n=0$,
$f_{2}, f_{1}, f_{0}+t_{1}$ do
not exist.
Therefore, for non-negative integers $0,1,2$,
$5, f_{\Delta}$ does not exist.