Solution

Answer: \(1344\)

Observe that \(1344\) is a Norwegian number as \(6\), \(672\) and \(1344\) are three distinct divisors of \(1344\) and \(6+672+1344=2022\). It remains to show that this is the smallest such number.

Assume for contradiction that \(N<1344\) is Norwegian and let \(N/a\), \(N/b\) and \(N/c\) be the three distinct divisors of \(N\), with \(a<b<c\). Then

\[2022=N\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)<1344\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\]

and so

\[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>\frac{2022}{1344}=\frac{337}{224}=\frac{3}{2}+\frac{1}{224}\]

If \(a>1\) then

\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}<\frac{3}{2}\]

so it must be the case that \(a=1\). Similarly, it must hold that \(b<4\) since otherwise

\[1+\frac{1}{b}+\frac{1}{c} \leqslant 1+\frac{1}{4}+\frac{1}{5}<\frac{3}{2}\]

This leaves two cases to check, \(b=2\) and \(b=3\). **Case** \(b=3\). Then

\[\frac{1}{c}>\frac{3}{2}+\frac{1}{224}-1-\frac{1}{3}>\frac{1}{6}\]

so \(c=4\) or \(c=5\). If \(c=4\) then

\[2022=N\left(1+\frac{1}{3}+\frac{1}{4}\right)=\frac{19}{12} N\]

but this is impossible as \(19 \nmid 2022\). If \(c=5\) then

\[2022=N\left(1+\frac{1}{3}+\frac{1}{5}\right)=\frac{23}{15} N\]

which again is impossible, as \(23 \nmid 2022\). **Case** \(b=2\). Note that \(c<224\) since

\[\frac{1}{c}>\frac{3}{2}+\frac{1}{224}-1-\frac{1}{2}=\frac{1}{224}\]

It holds that

\[2022=N\left(1+\frac{1}{2}+\frac{1}{c}\right)=\frac{3 c+2}{2 c} N \Rightarrow(3 c+2) N=4044 c\]

Since \((c, 3 c-2)=(c, 2) \in{1,2}\), then \(3 c+2 \mid 8088=2^{3} \cdot 3 \cdot 337\) which implies that \(3 c+2 \mid 2^{3} \cdot 337\). But since \(3c+2 \geqslant 3 \cdot 3+2>8=2^{3}\) and \(3c+2 \neq 337\), then it must hold that \(3c+2 \geqslant 2 \cdot 337\), contradicting \(c<224\).