Series of natural numbers which has all same digits 
For which $x$ exists sum $1 + 2 + 3 + \ldots + n$, where $n > 3$, which has notation $xxx\ldots x$?

In other words, I am looking for a sum of natural numbers which gives a result which has all same digits, e.g. $5555555$ or $22222222222222222$, or whatever else.
How to find this series and how to proof it for all possible series?
I have found the following examples:


*

*$1 + 2 + 3 + \ldots + 10 = 55$

*$1 + 2 + 3 + \ldots + 11 = 66$

*$1 + 2 + 3 + \ldots + 36 = 666$

And what about next series which ends with (different) repeating $x$? Hot to find and proof it?
 A: I found that it is known that the only triangular numbers (i.e. sums $1+2+\cdots+n$) which are also "repdigits" (i.e. all the same digit in base 10) are, as listed in the O.E.I.S as sequence  A045914 those in the list
$$0,\ 1,\ 3,\ 6,\ 55,\ 66,\ 666.$$
As you see they are including $0$ (the "empty sum" of digits, or the $0^{th}$ triangular number.) They also include the single digit numbers $1,3,6$ since technically they are repdigits of only one digit, and are also triangular. Thus it is known that the three you mention are the only ones of length 2 or more.
In a related article/list  A213516 of triangular numbers with at most two different digits, in the discussion they refer definitely to the fact that the above list is complete for the one digit triangular numbers. Also at the first reference A045914 there is a citation of a paper by D. Ballew and R. Wagner which appeared in J. Rec. Math. Vol 8 (2) p 96, year 1975-76, titled "Repdigit Triangular Numbers". I don't have access to that, but perhaps in that paper they have a proof that the above list is complete.
A: This is a partial answer. I can't access to the article of Ballew and Weger, so I'm trying to prove this myself.
Let's say that $9n(n+1)=2dK_r$ where $K_r=10^r-1$ for $r\ge 2$.
Some partial results:


*

*The digit $d$ is $1$, $5$ or $6$. Indeed, we know that $9+8dK_r$ must be a perfect square, so putting $d\in\{2,3,4,7,8,9\}$ and taking mod ${100}$ we get that 
$$9+8dK_r\equiv9-8d\in\{1,93,85,77,69,61,53,45,37\}\pmod{100}$$
There are only three quadratic residues mod $100$ in this list, namely $1$, $69$ and $61$, corresponding to $d=1,5,6$.

*If $d=5$ then $r$ is even. Because if $r=2t+1$ for some $t\ge 1$, we get that
$$9+40(10^{2t+1}-1)=4\cdot100^{t+1}-31$$
should be a perfect square, but since
$$(2\cdot10^{t+1}-1)^2<4\cdot100^{t+1}-31<(2\cdot10^{t+1})^2$$
it can not.
Some dead ends


*

*Taking mod $5^r$ for greater values of $r$ seems to be useless, because $9-8d=1,-31,-39$ are quadratic residues mod $5^r$ for every $r$.

*The suggestion of @Peter about comparing the fractional parts of $A=\sqrt{9n(9n+1)}$ and $B=\sqrt{2dK_r}$, doesn't seem to lead anywhere. The fractional part of $B$ is difficult to control. Nevertheless, I have computed some values for $d=1,6$ and the distribution of the fractional parts of $\sqrt{2dK_r}$ does not seem entirely random.
A: The sum of the first n digits is given by $\frac{n(n+1)}{2}$ and this must add up to repdigits in other words let the repeating digit be $x$ where $ x\in \{ 1,2,3,4,5,6,7,8,9,0\} $ then 
$$ \frac{n(n+1)}{2}=\underbrace{11..11}_{\text{k digits}}\times x$$
$$ n^2+n-\underbrace{22..22}_{\text{k digits}}\times x=0$$
$$\Delta = 1+\underbrace{88..88}_{\text{k digits}}\times x$$
Since $n$ must be a natural number $\Delta$ must be a perfect square. Therefor we are trying to find all the solutions to 
$$x=\frac{m^2-1}{\underbrace{88..88}_{\text{k digits}}} \text{ where } \{m\in Z|m>1\}$$
$\underbrace{88..88}_{\text{k digits}}$ can also be represented by $\frac{8}{9}(10^k-1)$, but I cannot tell if there is an arithmetic solution. With numeric methods I've gone up to 9 digits with no additional solutions. So regarding the upper bound what I can find is:
$$x=\frac{9(m^2-1)}{2^3(10^k-1)} <10$$
Using the quadratic formula $n=\frac{m-1}{2}$ simplifying to
$$x=\frac{9n(n+1)}{2(10^k-1)} <10$$ but i cannot yet find an upper bound for $m$, $n$ or $k$ using this.
A: I tried to prove this myself and found a quite elementary approach towards this problem which worked out to be really well.
Firstly, we know that,
$\sum_{i=1}^{n}i = \frac{n(n+1)}{2}$
Now,
$n(n+1) = 2x(1\cdots1)$
This is quadratic in $n$, so, discriminant of the above expression is:
$1 + (8\cdots8)x$
This has to be an odd perfect square for $n$ to be an integer. 
Rearranging the above expression,
$(8\cdots8)x = (m-1)(m+1), m \in \mathbb N$
Now, because $x$ only runs from $0$ to $9$, we can check the validity case by case.
Now, here’s the main idea. Clearly, $(8\cdots8)$ is even and equality of the form $(m-1)(m+1)$ is attained only when both the factors are even because $m$ = arithmetic mean of the two chosen factors. $(8\cdots8)x$ has only a handful of even factors (You can easily see why). 
Now because $(m+1) - (m-1) = 2$, the two factors chosen should also have a difference of two (because arithmetic mean of the two numbers is equidistant from both the numbers and here, that distance is just $1$) 
Let me do some cases for you:
1) $x = 1$,
$(8\cdots8) = 2(4\cdots4) = 4(2\cdots2)$,
As you can see, when the number of $8$s is 1,
we get $8 = 2\times4$ and this gives us $1$ as one of our answers and you can easily see in terms of the difference between the factors(which isincreasing with digits) that equality is never attained after this point.
2)$x=3$,
$3(8\cdots8) = 6(4\cdots4) = 12(2\cdots2)$,
As you can see, when the number of $8$s is 1,
we get $24 = 6\times4$ and this gives us $3$ as one of our answers and you can easily see in terms of the difference (which is increasing with digits) between the factors that equality is never attained after this point.
QED
