Cesenatico

Solution:

The possible values of $p(1)$ are $1, 1999, 3.996.001, 7.992.001$.

To justify these results, let us begin by proving that
- $a$ can only be $1$ or $1999$;
- $b$ is a positive divisor of $2001$;
- $b-a$ equals $\pm 1$ or $\pm 2$.

To prove these statements, observe that from $p(0)=0$ it follows that the constant term of $p(x)$ is zero, so we can write
$$p(x) = x q(x)$$
for some polynomial $q(x)$ with integer coefficients. Substituting $x=a$ into this relation we get $a q(a) = 1999$ and thus, since $a$ and $q(a)$ are integers and $1999$ is prime, the first statement is proved. The second statement is proved analogously by setting $x=b$.

For the third statement, observe that by Ruffini's theorem it follows that
$$p(x) = (x-a) r(x) + p(a) = (x-a) r(x) + 1999$$
for some polynomial $r(x)$ with integer coefficients. Substituting $x=b$ into this relation we get $(b-a) r(b) = 2$, that is, $b-a$ is a divisor of $2$, which is equivalent to the third statement.

Since $2001 = 3 \cdot 23 \cdot 29$ it is easy to check that the only positive values of $a$ and $b$ satisfying the three conditions are $a=1, b=3$, or $a=1999, b=2001$.

In the first case $p(1) = p(a) = 1999$. It remains to verify that there actually exist polynomials satisfying the conditions required in the statement with $a=1$ and $b=3$: for example the second-degree polynomial $-666 x^{2} + 2665 x$.

$$p(x) - x = x(x-1999)(x-2001) s(x)$$
for some polynomial $s(x)$ with integer coefficients. Substituting $x=1$ we then get that
$$p(1) = 1 + 1998 \cdot 2000 \cdot s(1)$$
Since by hypothesis we have $0 \leq p(1) \leq 10^{7}$, the only possible values for $s(1)$ are $0, 1, 2$, to which correspond $p(1) = 1, 3.996.001, 7.992.001$.