So far, I have come up with Cr being reduced from +6 to +3 and I being oxidized from -1 to 0.
K2Cr2O7(aq) +HI(aq) -> KI(aq)+ CrI3(aq) +I2(aq) +H2O(aq)
+1 +6 -2 +1 -1 +1 -1 +3 -1 0 +1 -2
Till this point you were on the right track.
After that you have gone wrong.
Think like this.
How many Cr atoms are reduced in K2Cr2O7? What is the total no. of electrons involved in this?
So how many electrons should be involved in oxidation of I- ? How many electrons are involved in oxidation of 1 mole of I- to I?
Then the method should automatically strike you