jmc

By the Chinese Remainder Theorem, it suffices to find the remainders when $N$ is divided by $4$ and $9$. The last two digits of $N$ must be one of $33,37,73,$ or $77$; each of these leave a remainder of $1$ after division by $4$. By the divisibility property of $4$, it follows that $N\equiv 1(\mathrm{mod}4)$.

The sum of the digits of $N$ is equal to $13\times 7+17\times 3=142=15\times 9+7$. This leaves a remainder of $7$ after division by $9$, so it follows that $N\equiv 7(\mathrm{mod}9)$.

By the Chinese Remainder Theorem and inspection, it follows that $N\equiv 25(\mathrm{mod}36)$ satisfies the two congruences, and so $N$ leaves a remainder of $\boxed{25}$ upon division by $36$.