We know that $P(x)$ is an $n$-th degree polynomial, and that the function

$$
f(x)=P(x) \cdot \sum_{i=1}^{n}|x-i|
$$

is differentiable for all real $x$. Determine all polynomials with this property!
Let $k$ be any integer between 1 and $n$. The function

$$
Q(x)=P(x)|x-k|=f(x)-P(x) \sum_{1 \leq i \leq n}|x-i| \quad(i \neq k)
$$

is differentiable at $x=k$, since $f(x)$, $P(x)$, and the $|x-i|$ functions are all differentiable here for any $i \neq k$. If $x>k$, the difference quotient of this function is

$$
\frac{Q(x)-Q(k)}{x-k}=P(x)
$$

which approaches $P(k)$ as $x$ approaches $k$. If, however, $x<k$,

$$
\frac{Q(x)-Q(k)}{x-k}=-P(x)
$$

which approaches $-P(k)$ as $x$ approaches $k$.

Since $Q(x)$ is differentiable at $x=k$, these two limits must be equal, so $P(k)=0$. Since $k$ was any integer between 1 and $n$, we now know that the $n$-th degree polynomial $P(x)$ has $n$ distinct roots, so it must be of the form

$$
P(x)=a(x-1)(x-2) \ldots(x-n)
$$

where $a$ is any non-zero number. This, however, satisfies the requirements based on the above reasoning.