Edit: swiss gambit posted while i was typing

no.

in order to reach the position either black has made x moves and white x+1; so some body has made an odd number of moves.

The rooks have only one square they can move to (after a knight moves out) so the rook moves away and moves back w times wich is a=2x an even number of times.

if the knights are are on their original sqares: it requires a knight an even number of moves to move from one square to a square of the same color; which gives b, even

so if the knights are on original squares total moves are a1+a2+b1+b2

which is a sum of even numbers which is even.

if the knights have switched: a kight requires an odd number of moves to move from a square to a square of opposite color; which gives c, odd. but the other knigt must also have moved to a square of opposite color. total knight moves are c1+c2=d which is a sum of 2 odd numbers, so d is even.

so if the knights have swithed; total moves are a1+a2+d which is a sum of even numbers which gives an even number.

so neither player could have made an odd number of movers.