I assume that in condition (2) you meant to say that each monomial appearing in the $F$-support of $Red(f)$ is $\leq f$. (In other words, you should not assume reduction simply take monomials to monomials.)

Assuming this slight change, the answer to your question is yes there is such a function (and it is uniquely determined if you add the extra condition that scalars pull out, and monomials either are sent to themselves or sums of multiples of smaller monomials).

**Existence**: Consider an arbitrary monomial $m$. If $m$ occurs as the largest monomial (under your well-ordering) of some element $f\in I$, then we can write $f=m-\sum_{m'<m}\alpha_{m',m}m'$ for some constants $\alpha_{m',m}\in F$. Further, if some $m'<m$ is also a largest monomial of some element $g\in I$, then after subtracting $\alpha_{m',m}g$ from $f$, we can assume $\alpha_{m',m}=0$. After a finite number of iterations of this process (using the fact $\leq$ is a well-ordering) we may assume $\alpha_{m',m}=0$ whenever $m'$ is a "largest monomial" of some element in $I$. An easy argument now shows that this $f$ is uniquely determined (subject to this constraint), and we define $Red(m)=\sum_{m'<m}\alpha_{m',m}m'$. If $m$ is not the largest monomial of any element in $I$, set $Red(m)=m$. Finally, define $Red$ on an arbitrary element of $K[X_1,\ldots, X_n]$ by $F$-linearity on the monomials. (So your additional condition $Red(kf)=kRed(f)$ also holds!)

Condition (2) obviously holds, and condition (3) holds by linearity of the definition. It also isn't too hard to check that condition (1) holds. So this gives you the function you seek.

**Uniqueness**: (Very Brief Sketch) Show that $Red$ has to agree with the function defined above, by transfinite induction over the monomials.

Note 1: The answer to your question is no if you assume that reduction is also a multiplicative ring homomorphism. For example if $n=1$, $I=(X_1^2-X_1)$, and the monomials are well-ordered in the obvious way $1<X_1<X_1^2<\ldots$, then $Red(X_1)^2=X_1^2\neq X_1=Red(X_1^2)$. This shouldn't be surprising, as most reduction rules are *not* ring homomorphisms.

Note 2: One usually wants $Red$ to be "simplifying" so you don't get weird reductions like $Red(X_1)=X_1X_2$, and this requires some restrictions on your well-ordering (as hinted at by Avi Steiner).