, 2023

Observe $1344$ is a solution as $6 + 672 + 1344 = 2022$. Claim that $1344$ is the smallest one. Towards contradiction, assume $N < 1344$ is also so-last-year, then there exist $a < b < c$, such that
$$2022 = N \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) < 1344 \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$$
and so
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} > \frac{2022}{1344} = \frac{3}{2} + \frac{1}{224}$$
If $a > 1$ then
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{1}{2} + \frac{1}{3} + \frac{1}{4} < \frac{3}{2},$$
so we must have $a = 1$. Similarly, we must have $b < 4$ since
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{1}{1} + \frac{1}{4} + \frac{1}{5} < \frac{3}{2},$$
For the case $b = 3$, we have $c = 4, 5$, and $2022 = \frac{19}{12}N$ or $2022 = \frac{23}{15}N$ respectively, both are impossible since neither $12$ nor $15$ divides $2022$. So $a = 1, b = 2$. Note $c < 224$ since
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} > \frac{3}{2} + \frac{1}{224}.$$
and so
$$2022 = N \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{3c+2}{2c} N \right)$$
Easy to see that $\gcd(3c+2, c) \mid 2$ hence $3c+2 \mid 2022 \cdot 4 = 2^3 \cdot 337$. $3c+2 > 8$ because $c > b = 2$.
So we must have $3c+2 = 2 \cdot 337$, hence $c = 224$, contradicts to the assumption $c < 224$.