Partial answer
The last digit of any perfect square must be 0, 1, 4, 5, 6, 9. So we must have $n \in \{ 0, 1, 4, 5, 6, 9 \}$ (mod 10).
Recall that the sum of the first $n$ integers is $\frac{n(n+1)}{2}$, and every integer is congruent modulo 9 to the sum of its digits. Reading down the diagonal of the base-9 multiplication table, we see that perfect squares end in 0, 1, 4, or 7. So we must have $\frac{n(n+1)}{2} \in \{ 0, 1, 4, 7 \}$ (mod 9). Equivalently, $n \in \{0, 1, 4, 7, 8\}$ (mod 9).
Combining the mod 10 and mod 9 results, we get:
$$n \in \{0, 1, 4, 9, 10, 16, 19, 25, 26, 31, 34, 35, 36, 40, 44, 45, 46, 49, 54, 55, 61, 64, 70, 71, 76, 79, 80, 81, 85, 89 \}~ (\text{mod } 90)$$
Now, let's consider the last two digits of the number. All perfect squares end in 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, or 96. Of these, the only ones containing consecutive digits are 01, 56, and 89. But since $n = 1$ is explicitly ruled out by the question, and $n = 6$ ($\sqrt{123456} \approx 351.36306$) and $n = 9$ ($\sqrt{123456789} \approx 11111.111061$) fail to produce perfect squares, then the last two digits must be part of $n$ itself.
Combining the mod 90 and mod 100 results, we get:
$$n \in \{0, 1, 4, 9, 16, 25, 36, 44, 49, 61, 64, 76, 81, 89, 100, 109, 116, 121, 124, 125, 136, 144, 161, 169, 181, 184, 189, 196, 216, 224, 225, 229, 241, 244, 256, 261, 269, 289, 296, 301, 304, 316, 324, 325, 341, 349, 361, 364, 369, 376, 396, 400, 404, 409, 421, 424, 436, 441, 449, 469, 476, 481, 484, 496, 504, 521, 529, 541, 544, 549, 556, 576, 584, 589, 601, 604, 616, 621, 625, 629, 649, 656, 661, 664, 676, 684, 700, 701, 709, 721, 724, 729, 736, 756, 764, 769, 781, 784, 796, 800, 801, 809, 829, 836, 841, 844, 856, 864, 881, 889\}~ (\text{mod } 900)$$
This doesn't solve the problem, but it would reduce the effort of a brute-force search, since you only have to try 110 of every 900 consecutive integers. I'll see if there's a way to filter the list even further.