Finding multiples of 3 or 5 with MIPS assembly (Project Euler #1)
Your code is very clean and simplistic, which is a good thing. There is just one glaring problem that I see.

div
As you may already know, the div instruction is very slow instruction. I don't know precisely how slow, but on the x86 systems, the instruction has a latency of about 70.
A better solution would be to treat two registers as counters for the numbers 3 and 5. Here is what I mean by that:

Load register A with 3
Load register B with 5
Subtract 1 from A
Subtract 1 from B
If A == 0, increment the total by the iterator. Load A with 3 and go to step 3.
If B == 0, increment the total by the iterator. Load B with 5 and go to step 5.

By replacing these steps with the div instruction, your code will run significantly faster.

If you would like me to show you how your code would look with the above change, let me know in a comment.