of polynomials of the form $a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}$, where $a_{1}, a_{2}, \cdots, a_{n}$ are distinct positive integers not greater than $n$. Find all values of $n$ for which such a decomposition is possible.
12. Suppose the decomposition into $k$ polynomials is possible. The sum of coefficients of each polynomial $a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}$ equals $1+\cdots+$ $n=n(n+1) / 2$ while the sum of coefficients of $1976\left(x+x^{2}+\cdots+x^{n}\right)$ is $1976 n$. Hence we must have $1976 n=k n(n+1) / 2$, which reduces to $(n+1) \mid 3952=2^{4} \cdot 13 \cdot 19$. In other words, $n$ is of the form $n=2^{\alpha} 13^{\beta} 19^{\gamma}-1$, with $0 \leq \alpha \leq 4,0 \leq \beta \leq 1,0 \leq \gamma \leq 1$. We can immediately eliminate the values $n=0$ and $n=3951$ that correspond to $\alpha=\beta=\gamma=0$ and $\alpha=4, \beta=\gamma=1$.
We claim that all other values $n$ are permitted. There are two cases. $\alpha \leq 3$. In this case $k=3952 /(n+1)$ is even. The simple choice of the polynomials $P=x+2 x^{2}+\cdots+n x^{n}$ and $P^{r}=n x+(n-1) x^{2}+\cdots+x^{n}$ suffices, since $k\left(P+P^{\prime}\right) / 2=1976\left(x+x^{2}+\cdots+x^{n}\right)$.
$\alpha=4$. Then $k$ is odd. Consider $(k-3) / 2$ pairs $\left(P, P^{\prime}\right)$ of the former case and
$$
\begin{aligned}
P_{1}= & {\left[n x+(n-1) x^{3}+\cdots+\frac{n+1}{2} x^{n}\right] } \\
& +\left[\frac{n-1}{2} x^{2}+\frac{n-3}{2} x^{4}+\cdots+x^{n-1}\right] ; \\
P_{2}= & {\left[\frac{n-1}{2} x+\frac{n-1}{2} x^{3}+\cdots+x^{n}\right] } \\
& +\left[n x^{2}+(n-1) x^{4}+\cdots+\frac{n+1}{2} x^{n-1}\right] .
\end{aligned}
$$

Then $P+P_{1}+P_{2}=3(n+1)\left(x+x^{2}+\cdots+x^{n}\right) / 2$ and therefore
$$
(k-3)\left(P+P^{\prime}\right) / 2+\left(P+P_{1}+P_{2}\right)=1976\left(x+x^{2}+\cdots+x^{n}\right) .
$$

It follows that the desired decomposition is possible if and only if $1<n<$ 3951 and $n+1 \mid 2-1976$.