The general form of the conjecture is false, due to the (often forgotten?) fact that dependent variables can be uncorrelated (i.e. zero covariance). This makes the RHS $=0$ and the rest is not difficult. Here is a simple counter-example:
$n=2$ and $X_1, X_2$ i.i.d. uniform $\in \{-1, 0, 1\}$ with prob $\frac13$ for each choice.
$Y_i = |X_i|$
$E[X_i Y_i] = \frac13 ( -1 + 0 + 1) = 0 = E[X_i]E[Y_i]$ so $Cov(X_i, Y_i) = 0$ i.e. they are uncorrelated. Thus RHS $=0$ because each term in the sum is either independent or uncorrelated.
Meanwhile these tables show $\max X_i$ (top table) and $\max Y_i$ (bottom table) for each of the $9$ equi-probable choices of $(X_1, X_2)$:
maxX -1 0 1
-1 -1 0 1
0 0 0 1
1 1 1 1
maxY -1 0 1
-1 1 1 1
0 1 0 1
1 1 1 1
From these tables:
$E[\max X_i] = \frac19 (5 - 1) = \frac49$
$E[\max Y_i] = \frac89$
$E[\max X_i \cdot \max Y_i] = E[\max X_i] = \frac49$
so $Cov(\max X_i, \max Y_i) = \frac49 - \frac49 \frac89 = \frac{4}{81} > 0 = $ RHS.
Further guess: I think adding "each variable positively correlated with each other" won't help, because a very small perturbation to above can (my guess) make the positive covariances on the RHS arbitrarily small, not enough to overwhelm the $\frac{4}{81}$ covariance on the LHS.
UPDATE: Here is a coin-based counter-example. The trick we now exploit is that pairwise independence does not equal mutual independence. And pairwise independence is sufficient to make RHS $= 0$.
$n = 2$ and $X_1, X_2$ i.i.d. uniform $\in \{0,1\}$ with prob $\frac12$ for each choice.
$Y_1 = Y_2 = \max Y_j = (X_1 \neq X_2)$, i.e. $(X_1 ~~~\text{xor}~~~ X_2)$
As is well known, $\forall (i,j) \in \{1,2\}^2: X_i, Y_j$ are (pairwise) independent, so $Cov(X_i, Y_j) = 0$ and RHS $=0$.
Here are all four possibilities:
X1 X2 maxX Y1=Y2=maxY
0 0 0 0
0 1 1 1
1 0 1 1
1 1 1 0
$E[\max X_i] = \frac34$
$E[\max Y_i] = \frac12$
$E[\max X_i \cdot \max Y_i] = \frac12$
$Cov(\max X_i, \max Y_i) = \frac12 - \frac12 \frac34 = \frac18 > 0 =$ RHS.