Solution:

If the leading digit is $a,$ the digits are $a,$ $a - 1,$ $a - 2,$ $a - 3$ with $a = 3, 4, \ldots, 9,$ so $$n = 1000a + 100(a-1) + 10(a-2) + (a-3) = 1111a - 123.$$

Since $1111 = 30 \cdot 37 + 1$ and $123 = 3 \cdot 37 + 12,$ we get $n \equiv a - 12 \equiv a + 25 \pmod{37}.$ For $a = 3, \ldots, 9$ the values $a + 25$ run through $28, 29, \ldots, 34,$ each already less than $37,$ so these are exactly the seven possible remainders.

Their sum is $28 + 29 + \cdots + 34 = 7 \cdot 31 = 217.$