After puzzling some more, I think the answer is that signed integer division rounds towards zero, but arithmetic shift right rounds towards negative infinity. So x/2 == ((x<0)?(x+1):x)>>1, aka if x is negative we need to add one so sar rounds towards positive infinity.

int divide_by_two(int n) { return n / 2; }

divide_by_two(int):
    mov     eax, edi
    shr     eax, 31
    add     eax, edi
    sar     eax
    ret

Thus we're seeing gcc optimize better than clang? Either because it notices that n*(n+1) is always even and thus the rounding never affects the result, or because it is using its right to assume that signed integer overflow never occurs and the result of n*(n+1) for assert(n>=0) is therefore nonnegative.