let g(x) = f(x) - x then the problem can be rewriten as g(g(x)+x) + g(x) = 0 and g(x) + g(y) >= 0 for all even x and odd y. Let A be the even number for which g(A) is minimal and B be the odd number for which g(B) is minimal (they exxist because if g is not bounded from below then it is easy to find contradiction to g(x)+g(y)>=0 for x,y with different parity). we have g(A) + g(B) >= 0 so one of them is at least 0, I will assume it is g(A) but the same proof will apply in the other case. if g(B) >= 0 then g(x)>=0 for all x then 0 = g(x) +g(g(x)+x)>=g(x) so g(x)<=0 for all x so g(x) = 0 for all x If g(B) < 0 then for all even x we have we have 0 =g(x) +g(g(x)+x) >= g(A) + g(B) >= 0 meaning all inequalities are actually equalities so g(x) = g(A) for all even x, g(g(x) + x) = g(B) for all even x but because g(x) = g(A) for all even x we deduce g(g(A) + x) = g(B) for all even x. it cannot be that g(A) is even because then we will get that g(g(A) + x) = g(A) for all even x as the argument is even and so we get g(B) = g(A) and because we assumed g(B) < 0 this would mean g(A) + g(B) < 0 contradiction. Because g(A) is odd we get that the g(g(A) + x) = g(B) for all even x just means g(y) = g(B) for all odd y. Now that we know that all even numbers are mapped to a single output and the same goes for all odd values it is easy to finish the rest.

Mikeglicker wrote: it is easy to finish the rest. $$ \boxed{ f(x) = \left\{\begin{matrix} x + a & x \text{ even}\\ x - a & x \text{ odd} \end{matrix}\right. \quad\text{for any }a = 0 \text{ or odd} } $$