A 3-digit number with equal digits is divisible by $111$ and hence by $37$ because $111 = 37 \cdot 3$. By hypothesis such a number is $1 + 2 + \cdots + (N-1) = \frac{1}{2}N(N-1)$, so $N(N-1)$ is divisible by $37$. Hence $N = 37k$ or $N = 37k+1$ for some integer $k \ge 1$. If $k \ge 2$ then $N \ge 74$, $\frac{1}{2}N(N-1) > 1000$ and $\frac{1}{2}N(N-1)$ is not a 3-digit number. Therefore $k=1$, i.e. $N = 37$ or $N = 38$. Since $1 + 2 + \cdots + 36 = 666$, $1 + \cdots + 37 = 703$, only $N = 37$ is a solution.