Thx for your contribution.Have to think about it a bit more...

@PietroMajer : nice idea !

@Jack L : for me, a decreasing function means : if $x\leq y$ then $F(x) \geq F(y)$.

Well $F$ is decreasing so, precisely, it switches the order.

Nope. $x^2\leq x^0$ because $(1,...1)$ is the max of the whole set but there is no need for $x^2$ to be smaller than $x^1$

@Pietro Majer : can you develop ? Because $F(x)\leq x $ is not necessarly satisfied.

@Benjamin Steinberg : decreasing in the standard sense for the order product : if $x\leq y$ then $F(x)\geq F(y)$.

Some others comments : if you remove one of the two hypothesis, the result is wrong even for $N=2$. Moreover, I have an algorithm that produces a lot of examples of such $F$ for which the proposition was always satisified.

No, the order is the product order which is only a partial order: $$(x_1,x_2,\cdots,x_N)\leq(y_1,y_2,\cdots,y_N)$$ if and only $$x_i\leq y_i$$ for all i=1..N. Moreover, LSpice is correct. Thanks for your interest.

It is for N=2. Just check all the possible $F$

Yes, you are right. I corrected.

Well, I am pretty convinced that this is true : I have an implemented an algorithm that computes the correct sequence of $\Delta_i$. This algorithm never failed. The most strange fact, at least to me, is the uniqueness. Notice that if, for instance, the integer $n_i$ is bigger than $4$ then the only possibility is $\Delta_i=0$. Thus, the problem reduces to small values of $n_i$. One last comment : this question comes from some consideration in geometry of germ of complex curve in the complex plane.