The only two such a are $a = 5$ and $a = 6$. considering $ \frac{n(n+1)}{2} (\text{mod } 100) $ we see only $ a=1,5,6 $ can work as the last two digits must be the same and note that $n=10$ and $n=11$ work for $a= 5$ and $a = 6$ respectively. Now suppose it can be done for $ a = 1 $ i.e. for some $ n $, $ \frac{n(n+1)}{2} = \underbrace{1 \cdots 1}_k = \frac{10^k-1}{9} $ for some $ k $ and note that $ k>1 $ as $ n \geq 4 $. Then we must have $ 2^{k+1} 5^k = 9n^2+9n+2 = (3n+1)(3n+2) $ which are co-prime so one is $ 2^{k+1} $ and the other is $ 5^k $ but for $ k>1 $ their difference is at leasts 2 which is a contradiction so it cant be done for $ a = 1 $

Suppose $\overbrace{aaa\ldots a}^{m\ \text{times}}=\frac{n(n+1)}{2}$. The number in the $LHS$ is equal to $a\cdot \overbrace{111\ldots 1}^{m\ \text{times}}=a\cdot \frac{10^{m+1}-1}{9}$. Hence we aim to solve the equation $\frac{n(n+1)}{2}=a\cdot \frac{10^{m+1}-1}{9} \implies 9n^2+9n=2a10^{m+1}-2a$. The last equation becomes $n^2+n=2a \pmod{10}$. The only possible values of $n^2+n \pmod{10}$ are $0,2$ and $6$. Hence $a$ can only equal $0,1,3,5,6,8 \pmod{10}$ i.e. $a$ can only equal $0,1,3,5,6,8$ since $a<10$. Case 1 $a=0$ Trivially there are no solutions. Case 1 $a=1$ So $9n^2+9n+2=2\cdot 10^{m+1}$ which cannot happen $\pmod{25}$ - note $m>0$. Case 3 $a=3$ Then $9n^2+9n=6\cdot 10^{m+1}-6 \implies 3n^2+3n+2=2\cdot 10^{m+1}=0 \pmod{25}$. Again this cannot happen. Case 4 $a=5$ $T_{10}=55$ so $a=5$ is a solution. Case 5 $a=10$ $T_{11}=66$ so $a=6$ is a solution. Case 6 $a=8$ Then $9n^2+9n+16=16\cdot 10^{m+1}$, a contradiction $\pmod{25}$. Thus only solutions are $a=6$ and $a=5$.