Assume that $\frac{m(m+1)}{2}=r$ and 111..(k 1's).|r We can show that $k<2$ So $a=(1,3,6)$

For $a=1,3,6,5$ pick $n=1,2,3,10$, respectively. If $\frac{n(n+1)}{2} = a\frac{10^k - 1}{9}$, then $(2n+1)^2 = \frac{8a}{9}(10^k - 1) + 1$ and now modulo $5$ gives that $8a+1 \equiv 0,1,4 \pmod 5$, which rules out $a=2,4,7,9$. If $a=8$ then mod $5$ gives that $(2n+1)^2$ is divisible by $5$, so $64 \cdot (11\cdots 1) + 1$ must be divisible by $25$, but for $k\geq 2$ (rule out $k=1$ by direct verification) it is congruent to $64 \cdot 11 + 1 \equiv 5 \pmod {25}$, contradiction.