1. The original proposition can be rewritten as: For any polynomial $R(x)$ of degree not exceeding $n$, there exist monomials $a x^{k}$ and $b x^{l}$ (where $a, b$ are real numbers, and integers $k, l$ satisfy $0 \leqslant k < l \leqslant n$), such that $R(x) + a x^{k} + b x^{l}$ always has no roots.

If $n \geqslant 3$ is odd, for $R(x) = x^{n} + x$, to avoid obtaining an odd-degree polynomial, which would have roots, we must take $b x^{i} = -x^{n}$.
When $k \geqslant 1$,
$x = 0 \Rightarrow R(x) + a x^{k} + b x^{l} = 0$;
When $k = 0$, it is easy to see that $x + a$ always has a root.
If $n = 1$, we always have $R(x) + b x + a \equiv 1$, which is not 0.
If $n$ is even, then we can add a monomial of degree $n$ to make the leading coefficient positive, and the new polynomial will have a lower bound $M$. Then, by adding a constant $1 - M$, the minimum value of the new polynomial will be 1, i.e., it has no roots.
In summary, the $n$ that makes the proposition true are even numbers or 1.
1. Find all positive integers $n$ such that for any two $n$-degree polynomials $P(x), Q(x)$, there exist monomials $a x^{k}, b x^{l}$ (where $a, b$ are real numbers, and integers $k, l$ satisfy $0 \leqslant k, l \leqslant n$), such that the graphs of $P(x)+a x^{k}, Q(x)+b x^{l}$ have no intersection points.