Answer

Solution

1. 1

   Average $k_i=(5n-4)/n=5-4/n$. So all $k_i\le 5n-4$. Also we need $\sum 1/k_i=1$.

2. 2

   Try $n=4$: $\sum k_i=16,\;\sum 1/k_i=1$. The classical Egyptian fraction $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}=1$. Check sum: $2+3+7+42=54\ne 16$. Try $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=1$, sum $=24\ne 16$. Try $\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+\frac{1}{12}=1$, sum $3+3+4+12=22$.

3. 3

   Try $\{2,4,5,5\}$: $\frac12+\frac14+\frac15+\frac15=\frac{10+5+4+4}{20}=\frac{23}{20}\ne 1$.

4. 4

   Try $n=3$: $\sum k_i=11,\;\sum 1/k_i=1$. $\frac12+\frac13+\frac16=1$, sum $=2+3+6=11$. Works!

5. 5

   Try $n=4$: $\sum k_i=16,\;\sum 1/k_i=1$. By AM-HM, $\frac{\sum k_i}{n}\ge \frac{n}{\sum 1/k_i}=\frac{4}{1}=4$, so $\sum k_i\ge 16$ with equality iff all $k_i$ equal. Equality forces $k_i=4$, but then $\sum 1/k_i=1$ ✓ and $\sum k_i=16$ ✓. So $\{4,4,4,4\}$ works for $n=4$.

6. 6

   Try $n=5$: $\sum k_i=21,\sum 1/k_i=1$. AM-HM: $\sum k_i\ge n^2/\sum(1/k_i)=25$. But $21<25$, contradiction. So $n=5$ impossible.

7. 7

   Maximum $n=4$.

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