Problem 1256

### Case 1: All elements of $T$ are irrational numbers

1. **Assume the form of the set $T$**:
Since $p(T) - a$ is an odd integer for any $a \in T$, we can infer that the difference between any two elements of $T$ must be an even integer. Therefore, we assume the set $T$ is of the form:
$$T = \{x, x+2, x+4, \ldots, x+2 \cdot 2020\}$$
where $x$ is an irrational number.

2. Existence of a root:
Consider the polynomial equation formed by the product of the elements of $T$:
$$x(x+2)(x+4) \cdots (x+4040) = x + 1$$
Since the degree of the polynomial on the left-hand side is odd (2021), it must have at least one real root. By the Rational Root Theorem, if this root were rational, it would have to be $\pm 1$. However, due to the size and nature of the polynomial, it is clear that $\pm 1$ cannot be roots. Therefore, there exists an irrational root $x_0$.

3. Conclusion for irrational numbers:
The set $\{x_0, x_0+2, \ldots, x_0+4040\}$ satisfies the condition that $p(T) - a$ is an odd integer for any $a \in T$.

### Case 2: At least one element of $T$ is a rational number

1. **Assume the form of the set $T$**:
By the same logic as before, the elements of $T$ must differ by an even integer. Therefore, we assume the set $T$ is of the form:
$$T = \{x, x+2, x+4, \ldots, x+2 \cdot 2020\}$$
where $x$ is a rational number.

2. Valuation argument:
Let $x = \frac{p}{q}$ be a rational number in lowest terms. Consider the $p$-adic valuation $v_p$. If $v_p(x) = -a < 0$, then:
$$v_p(p(T)) = v_p\left(\frac{p}{q} \cdot \frac{p+2q}{q} \cdot \frac{p+4q}{q} \cdots \frac{p+4040q}{q}\right) = -2021a$$
However, for $p(T) - a$ to be an odd integer, $v_p(p(T) - a)$ must also be $-a$. This implies $a = 0$, meaning the rational numbers have no denominator and are integers.

3. Modulo 2 argument:
Since the elements are integers differing by an even amount, consider the product modulo 2. If all elements are odd, then:
$$p(T) \equiv 1 \cdot 1 \cdot \ldots \cdot 1 \equiv 1 \pmod{2}$$
and:
$$p(T) - a \equiv 1 - 1 \equiv 0 \pmod{2}$$
If all elements are even, then:
$$p(T) \equiv 0 \cdot 0 \cdot \ldots \cdot 0 \equiv 0 \pmod{2}$$
and:
$$p(T) - a \equiv 0 - 0 \equiv 0 \pmod{2}$$
In both cases, $p(T) - a$ cannot be an odd integer, leading to a contradiction.

4. Conclusion for rational numbers:
It is impossible to have at least one rational number in the set $T$ while satisfying the given condition.

$\blacksquare$