The answer: Yes! Take P(x)=x^2(x+1)^2(x+2)^2...(x+2022)^2 And take Q(x)=(x+1)^2(x+2)^2...(x+2024)^2. Then it's easy to check that these polynomials satisfy conditions of the problem.

We will prove that for all positive integers n there are polynomials $P, Q$ with real coefficients, such that $P(P(x))\cdot Q(Q(x))$ has exactly $n$ distinct real roots and $P(Q(x)) \cdot Q(P(x))$ has exactly $n+1$ distinct real roots. Will construct $P$ and $Q$. We take $0<p_1<p_2<\ldots<p_{n-1}$ $n-1$ distinct positive real numbers and $0<q_1<q_2<\ldots<q_{n+1}$ other $n+1$ distinct positive real numbers .Let $P(x)=x^2(x+p_1)^2(x+p_2)^2...(x+p_{n-1})^2$ and $Q(x)=(x+q_1)^2(x+q_2)^2...(x+q_{n+1})^2$. So $P(P(x))=P(x)^2\cdot R(x)$ where $R$ is a polynomial with real coefficients and with $0$ real roots and $Q(Q(x))=T(x)$ where $T$ is a polynomial with real coefficients and with $0$ real roots. So $P(P(x))\cdot Q(Q(x))$ has exactly $n$ real roots. $P(Q(x))=Q(x)^2\cdot U(x) $ where $U$ is a polynomial with real coefficients and with $0$ real roots and $Q(P(x))=V(x)$ where $V$ is a polynomial with real coefficients and with $0$ real roots.So $P(Q(x)) \cdot Q(P(x))$ has exactly $n+1$ real roots, so we are done.

Groupsolved with GeorgeRP (BGRB4) and Lazar Delyanov Todorov (BGRB2) during the BMO training camp. The answer is yes. We boldly choose $P(x) = x^2+1$ and pick \[Q(x) = -tx(x-x_1)(x-x_2)\ldots(x-x_{1009})(x-x_{1010})^2(x-x_{1011})\ldots (x-x_{2022}).\]Here, $t$ is some positive real number to be chosen later, and $0<x_1<x_2<\ldots<x_{2022}$. The idea is that $P(P(x))=0$ has no roots, so we want $Q(Q(x))=0$ to have exactly $2023$ distinct real roots. We can achieve this by picking a suitable $t$. Indeed, let $\sup \frac{1}{t}Q(x) = q$. The polynomial $Q_{1}(x) = \frac{1}{t}Q(x)$ has been specifically picked so that \[\lim_{x\to-\infty} Q_1(x)=-\infty = \lim_{x\to +\infty} Q_1(x)\]as $\deg Q_1 = 2024$ and the leading coefficient of $Q_1$ is $-1$. Therefore, $\sup Q_1$ exists. Now choosing $t>0$ such that $tq<x_1$ suffices. Indeed, if we consider the equation $Q(Q(x)) = 0$, this splits into $Q(x) = 0$, which has $2023$ distinct real roots, and equations of the type $Q(x)-x_i=0$, none of which has a solution as \[Q(x)-x_i=tQ_1(x)-x_i\leq tq-x_1<0.\]Therefore, $P(P(x))\cdot Q(Q(x)) = 0$ has exactly $2023$ distinct real roots, and we are left to show that $P(Q(x)) \cdot Q(P(x)) = 0$ can have exactly $2024$ roots. Clearly, $P(Q(x))=0$ has no roots, so we just require $Q(x^2+1)=0$ to have $2024$ roots. This is done by choosing $x_{1010}<1<x_{1011}$, which completes the solution as now only $Q(x^2+1)-x_{1010+i}=0$ have a solution for $i>0$, and these produce two distinct solutions each, so we're done.