Post History

Only if: suppose $N = 2^a m$ with $m$ odd and greater than $1$.

Because the odd numbers form a multiplicative group modulo $2^{a+1}$, there are $j_0$, $j_1$ such that $j_0 m \equiv 1 \pmod{2^{a+1}}$ and $j_1 m \equiv -1 \pmod{2^{a+1}}$. We have $(j_0 + j_1)m \equiv 0 \pmod{2^{a+1}}$ and since $m$ is odd, $j_0 + j_1 \equiv 0 \pmod{2^{a+1}}$. They're both odd, so identifying the equivalence classes with their representatives in the range $(0, 2^{a+1})$ we get $j_0 + j_1 = 2^{a+1}$, whence $\min(j_0, j_1) < 2^a$.

If $j_0 < 2^a$, consider $(j_0 m - 1)(j_0 m)$. It is divisible by $2^{a+1}$ and $m$, so $\Delta_{j_0 m - 1} \equiv 0 \pmod N$, but $j_0 m - 1 \not\equiv 0 \pmod N$.

If $j_1 < 2^a$, consider $(j_1 m)(j_1 m + 1)$. It is divisible by $2^{a+1}$ and $m$, so $\Delta_{j_1 m} \equiv 0 \pmod N$, but $j_1 m \not\equiv 0 \pmod N$.

Either way, a counting argument shows that some equivalence class modulo $N$ is not covered.

The assumption that $m$ is greater than $1$ is relevant because otherwise $j_0 = 1$ and we don't show that the equivalence class of $0$ is covered by multiple triangle numbers.