To look for a possible smaller n, we'll start by taking each prime factor 5, 13, and 31 in turn finding n such that the primes individually divide 3^n-2^n; and then combine those conditions to get the answer.
p=5: both exponentials have a cycle of length of 4. Starting at n=1:
2^n mod 5: 2, 4, 3, 1
3^n mod 5: 3, 4, 2, 1
These remainders are equal when n is even.
p=13: 2^n has a cycle of length 4 and 3^n has a cycle length of 12. Starting at n=1:
2^n mod 5: 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1
3^n mod 5: 3, 9, 1, 3, 9, 1, 3, 9, 1, 3, 9, 1
These remainders are equal when n is a multiple of 4.
p=31: 2^n has a cycle of length 5 and 3^n has a cycle length of 30. Starting at n=1:
2^n mod 5: 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1
3^n mod 5: 3, 9, 27, 19, 26, 16, 17, 20, 29, 25, 13, 8, 24, 10, 30, 28, 22, 4, 12, 5, 15, 14, 11, 2, 6, 18, 23, 7, 21, 1
These remainders are equal when n is a multiple of 30.
So now we need n such that it is even, a multiple of 4, and a multiple of 30. Then n is a multiple of 60, so the minimum value of a positive integer N such that: 3^N - 2^N is a positive integer multiple of 2015 is N=60.
Solution
