Modular quadratic equation question- Where did I go wrong? \begin{gather}
\frac{N^2+N}{2} \equiv 0 \pmod 4 \\
N^2+N\equiv 0\\
(N+\frac{1}{2})^2-\frac{1}{4}\equiv 0\\
4(N+\frac{1}{2})^2-1\equiv 0\\
2^2(N+\frac{1}{2})^2\equiv 1\\
(2N+1)^2\equiv 1 \\
2N+1 \equiv 1\\
2N=0\\
N=2,4,6,8...\\
2N+1 \equiv 3\\
2N=2\\
N=1,3,5,7...\\
\end{gather}
In other words I've shown $\forall N >0 \to \frac{N^2+N}{2} \equiv 0 \pmod 4$, which is false e.g. for $N=3$,  $\frac{N^2+N}{2}$ has a remainder of 2
I'd like a correct solution as well. Thanks.

What you did incorrectly is basically explained in Peter's answer, and several question comments. Note, though, the more general issue when manipulating congruences, since you are really dealing with integers, is that involving "fractions" of the form $\frac{p}{q}$ where $\gcd(p,q) = 1$, really means the division by $q$ is multiplying by its modulo multiplicative inverse, i.e., $q^{-1}$. However, only integers which are relatively prime to the modulus have inverses. Since $\gcd(4, 4) = 4 \neq 1$ and $\gcd(4, 2) = 2 \neq 1$, neither $2$ nor $4$ have an inverse in modulo $4$, so while working in that modulus you can't, in general, just "divide" by either value, e.g., $\frac{1}{2}$ and $\frac{1}{4}$ don't make sense.
As for how to correctly solve your congruence equation, note for any solution $N$, there's some integer $j$ where
$$\begin{equation}\begin{aligned}
& \frac{N^2 + N}{2} \equiv 0 \pmod{4} \iff \frac{N^2 + N}{2} = 4j \iff \\
& N^2 + N = 8j \iff N^2 + N \equiv 0 \pmod{8}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Due to $N$ and $N + 1$ being relatively prime to each other, with one being odd and one being even, then since $8 = 2^3$ has only prime factors of $2$, this then gives
$$\begin{equation}\begin{aligned}
& N(N + 1) \equiv 0 \pmod{8} \implies \\
& (N \equiv 0 \pmod{8}) \lor (N + 1 \equiv 0 \pmod{8}) \implies \\
& N \equiv 0, 7 \pmod{8}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
This means the allowed values for $N$ are $N = 8k$ and $N = 8k + 7$ for any $k \in \mathbb{Z}$.
