I wouldn't have answered something that looks so much like homework, but the "idiomatic" comment just begged for a demonstration that it's really not that far. First, a direct translation:
(define (sum-fibonacci-values)
(define n 4000000)
(define i2 1)
(define i1 1)
(define fibo 0)
(define temp 1)
(let loop ()
(when (< i2 n)
(set! temp (+ i1 i2))
(set! i1 i2)
(set! i2 temp)
(when (zero? (modulo i2 2)) (set! fibo (+ fibo i2)))
(loop)))
(write fibo))
Second, make it "idiomatic", by removing the redundant mutations, and instead just bind new values, and using a tail-recursive loop. Note that this code is still in direct correlation with the original:
(define (sum-fibonacci-values)
(define n 4000000)
(let loop ([i2 1] [i1 1] [fibo 0] [temp 1])
(if (< i2 n)
(let* ([temp (+ i1 i2)]
[i1 i2]
[i2 temp]
[fibo (if (zero? (modulo i2 2)) (+ fibo i2) fibo)])
(loop i2 i1 fibo temp))
fibo)))
Finally, now that the code is clearer, you can see that there are some redundancies. Here's a cleaned up version:
(define (sum-fibonacci-values)
(define n 4000000)
(let loop ([i2 1] [i1 1] [fibo 0])
(if (< i2 n)
(let ([i3 (+ i1 i2)])
(loop i3 i2 (if (zero? (modulo i3 2)) (+ fibo i3) fibo)))
fibo)))
Note that the same cleanup can be done on the Java code. (But that's really left as an exercise to the reader...)