3 For which positive integers $n$, do there exist positive integers $m$ and positive integers $a_{1}, a_{2}, \cdots, a_{m-1}$, such that $n=a_{1}\left(m-a_{1}\right)+a_{2}\left(m-a_{2}\right)+\cdots+a_{m-1}(m-$ $\left.a_{m-1}\right)$. Where $a_{1}, a_{2}, \cdots, a_{m-1}$ can be the same, and $1 \leqslant a_{i} \leqslant m-1$ $(i=1,2, \cdots, m-1)$.
Let $n \geqslant 12$ and $n$ does not satisfy the requirement.
Let $m$ be an even number, $m-2 l, b_{i}=\left|l-a_{i}\right|(i=1,2, \cdots, m-1)$, then
$$
0 \leqslant b_{i} \leqslant l-1(1 \leqslant i \leqslant m-1),
$$
II.
$$
\begin{array}{c}
a_{1}\left(m-a_{1}\right)+a_{2}\left(m-a_{2}\right)+\cdots+a_{m-1}\left(m-a_{m}\right) \\
=l^{2}(2 l-1)-b_{1}^{2}-b_{2}^{2}-\cdots-b_{2 l-1}^{2} .
\end{array}
$$

Let $l$ be the smallest positive integer such that $l^{2}(2 l-1) \geqslant n$.
Let $r=l^{2}(2 l-1)-n=(l-1)^{2} t+s$, where $t$, $s$ are non-negative integers, $0 \leqslant s(l-1)^{2}(2 l-3)$, so $r<7(l-1)^{2}$. Hence $t \leqslant 6$.

Let $k_{1}$ be a non-negative integer, $k_{1}^{2} \leqslant s<\left(k_{1}+1\right)^{2}$, then $\left\{1, k_{1}^{2} \leqslant s<(l-1)^{2}\right.$ implies $k_{1} \leqslant l-2$. Thus $0 \leqslant s-k_{1}^{2} \leqslant 2 l-4$.

If $s-k_{\uparrow}^{2} \leqslant 3$. Then $s$ can be expressed as the sum of no more than 4 squares, $4 \leqslant 2 l-1-t$, contradiction.

If $4 \leqslant s-k_{1}^{2} \leqslant 7$. Then $s$ can be expressed as the sum of no more than 5 squares. $5 \leqslant$ $2 l-1-t$. Contradiction.

If $s-k_{1}^{2} \geqslant 8$, then $0 \leqslant s-k_{1}^{2}-2^{2}-2^{2} \leqslant 2 l-12 \leqslant 2 l-1-$ $t-3 . s$ can be expressed as the sum of no more than $2 l-1-t$ squares, contradiction.
Therefore $l \leq 5$.
When $l=5$, $2 l-1=9,0 \leqslant r<6 l^{2}-8 l+3=113$.
Let $r=4^{2} t+s, 0 \leqslant s \leqslant 15$. Then $t \leqslant 7$.
If $t=7$. Then by $r \leqslant 112$ we know $s=0$. At this time, $r$ can be expressed as the sum of 7 $4^{2}$. Contradiction.
If $t=6$, it can be verified that: when $0 \leqslant s \leqslant 15, s \neq 7,15$, $s$ can be expressed as the sum of no more than 3 squares.

And $4^{2} \times 6+7=4^{2} \times 5+3^{2}+3^{2}+2^{2}+1^{2}, 4^{2} \times 6+15=$ $4^{2} \times 5+3^{2}+3^{2}+3^{2}+2^{2}$, contradiction.
When $l=4$, $2 l-1=7,0 \leqslant r<6 l^{2}-8 l+3=67$.
If $t \leqslant 4, s \neq 7$, then $s$ can be expressed as the sum of no more than 3 squares, $3 \leqslant 2 l-$ $1-t$, contradiction.

If $1 \leqslant t \leqslant 4, s=7$, then $r=3^{2}(t-1)+4 \times 2^{2}, t-1+4 \leqslant$ 7. Contradiction.
If $t=0, s=7$, then $r$ can be expressed as the sum of 7 $1^{2}$. Contradiction.
Therefore $t \geqslant 5$. Thus $r \geqslant 3^{2} t \geqslant 45, n=l^{2}(2 l-1)-r \leqslant 112-45$ $=67$.
When $l \leqslant 3$, $n \leqslant l^{2}(2 l-1) \leqslant 45$.
We only need to consider $n \leqslant 67$.
By $a_{1}\left(m-a_{1}\right)+\cdots+a_{m}\left(m-a_{m-1}\right) \geqslant(m-1)^{2}$. We only need to consider $m \leqslant 9$.
\begin{tabular}{|c|c|c|}
\hline$m$ & $a(m \quad a)$ possible values & $m \quad 1$ values of the form $u(m--a)$ (not exceeding 67) \\
\hline 9 & $8 \cdot 14 \cdot 18 \cdot 20$ & 64 \\
\hline 8 & $7 \cdot 12 \cdot 15 \cdot 16$ & $49 \cdot 54,57 \cdot 59,62,63,64,65 \cdot 66 \cdot 67$ \\
\hline 7 & t. 10.12 & \begin{tabular}{l}
\begin{tabular}{l}
$36 \cdot 40,42 \cdot 44 \cdot 46 \cdot 48 \cdot 50,52 \cdot 54,56,58,60$, \\
$62 \cdot 64,66$
\end{tabular}
\end{tabular} \\
\hline 6 & 5.8 .9 & \begin{tabular}{l}
\begin{tabular}{l}
$25,28,29,31,32,33,34,35,36,37,38,39 \cdot$ \\
$40,41 \cdot 42 \cdot 43,44,45$
\end{tabular}
\end{tabular} \\
\hline$\overline{5}$ & 4.6 & $16,18,20,22,24$ \\
\hline 4 & 3.4 & $9,10,11,12$ \\
\hline
\end{tabular}
\begin{tabular}{|c|c|c|}
\hline$m$ & $a(m-a)$ possible values & $m-1$ values of the form $a(m-a)$ (not exceeding 67) \\
\hline 3 & 2 & 4 \\
\hline 2 & 1 & 1 \\
\hline
\end{tabular}

The table shows that the $n$ that do not satisfy the requirement are: $2,3,5,6,7,8,13,14,15,17$,
19. 21 . $23,26,27,30,47.51,53.55 .61$, the rest of the $n$ satisfy the requirement.