There is an inequality between arithmetic and harmonic means of $n$ positive numbers:
$$\frac{k_1+\dots+k_n}{n} \geq \frac{n}{\frac{1}{k_1}+\dots \frac{1}{k_n}}$$
where $k_1>0, \dots, k_n>0$. Mathematica ( version $\geq$ 10.1) knows this relation for $n\leq 4$, e.g.
Simplify[Mean[#] >= HarmonicMean[#], Min[#] > 0] &[{k1, k2, k3, k4}]
True
It appears to be sufficient for our purpose, since from the assumptions we have:
$$\frac{k_1+\dots+k_n}{n} \geq n$$
because $\frac{1}{k_1}+\dots \frac{1}{k_n}=1$. On the other hand $k_1+\dots+k_n =5n-4$, therefore we get the following inequality:
Solve[ 5n - 4 >= n^2 && n > 0, n, Integers]
{{n -> 1}, {n -> 2}, {n -> 3}, {n -> 4}}
There are four possibilities: the first one yields an obvious solution $n=1$ and $k_1=1$. From the assumptions it is clear that, if $(k_1, \dots, k_n)$ is a solution than any permutation $(k_{\sigma(1)}, \dots, k_{\sigma(n)})$ is also a solution. Then for the sake of simplicity and a good performance we define a function yielding solutions ordered by $k_1\leq k_2 \leq \dots \leq k_n$:
sol[l_List] /; Length[l] > 0 :=
Solve[{ Mean @ l == 5 - 4/Length @ l, HarmonicMean @ l == Length @ l,
LessEqual @@ l, Min @ l > 0}, l, Integers]
we can find all solutions
sol /@ Table[Array[k, m], {m, 4}] // Column
{{k[1] -> 1}}
{}
{{k[1] -> 2, k[2] -> 3, k[3] -> 6}}
{{k[1] -> 4, k[2] -> 4, k[3] -> 4, k[4] -> 4}}
Reassuming, all possible solutions (up to permutations of $k_1, \dots ,k_n$) are:
$$ n=1,\quad k_1=1$$
$$ n=3,\quad k_1=2,\; k_2=3,\; k_3=6$$
$$ n=4, \quad k_1=k_2=k_3=k_4=4$$
k[c]
$\endgroup$ – Nasser Dec 14 '17 at 17:29