I think this problem is quite boring and so we have no solutions. But because this is in the <<Contests>> we should post one.

Well, what's the ideal solution here? Or is it just a boring casework? (writing $n$ in different forms considering that it has 16 divisors)

This problem is very easy.It's in Problem and solution around the world 1999-2000.

Mind writing your solution tkhtn please?

Sorry! This problem is in Balkan,Junio,2002. And here is solution: If$d_2>2$ then all divisorsof n are odd but $d_4+d_6$ is divisor of$n$ and it's even,contradiction],if$d_3=3$ then 6 is divisor of n$\to$ or $d_5=6$ or $d_6=6$or $d_4=6$. Assume $d_4=6$ then $d_2+d_4=8$ is divisor of$n \to 4$is divisor of$n \to d_4=4$,contradiction. Assume $d_5=6$ then $d_6=(d_2+d_4)d_6>d_6$,contradiction.......... And $N=2002$. sory because I have go to bed. See you agains.

HM . continue. I think this problem must be complete today. Assume $d_6=6$ then $d_4=4;d_5=5$.But $d_5=(d_2+d_4)d_6>d_6$,contradiction. $\to d_3>3\to$3 isn't a divisor of n. Because $(d_2+d_4)d_6$ is a divisor of n $\to (d_2+d_4)=d_4+2$ is divisor of n.if $d_4 +1$ is a divisor of n then we have three consecutive divisors $\to$ 3 is a divisor of n, contradiction $\to d_5=d_4+2$and $d_5=3k+1\to d_5={1,4,7,10,13 or 16}$.Obvious $d_5\ge 5$. Assume $d_5=7\to d_3=4,d_4=5\to 10,14$is divisors of n $\to d_7\le 10$. But $d_7=(d_2+d_4)d_6>(2+5)7=49$,contradiction. Other case similar and last:$d_5=13$ is true$\to d_4=11\to d_3={4 or 7}$. Now is very easy.

HM . continue. I think this problem must be complete today. Assume $d_6=6$ then $d_4=4;d_5=5$.But $d_5=(d_2+d_4)d_6>d_6$,contradiction. $\to d_3>3\to$3 isn't a divisor of n. Because $(d_2+d_4)d_6$ is a divisor of n $\to (d_2+d_4)=d_4+2$ is divisor of n.if $d_4 +1$ is a divisor of n then we have three consecutive divisors $\to$ 3 is a divisor of n, contradiction $\to d_5=d_4+2$and $d_5=3k+1\to d_5={1,4,7,10,13 or 16}$.Obvious $d_5\ge 5$. Assume $d_5=7\to d_3=4,d_4=5\to 10,14$is divisors of n $\to d_7\le 10$. But $d_7=(d_2+d_4)d_6>(2+5)7=49$,contradiction. Other case similar and last:$d_5=13$ is true$\to d_4=11\to d_3={4 or 7}$. Now is very easy.

Could you explain why this equality holds $d_5=(d_2+d_4)d_6$ because we only know that $d_{d_5}=(d_2+d_4)d_6$ BTW, do not answer with a word only.

Because $d_5=5$.

Is there another, shorter solution?

Ok so if all the divisors are odd, then $d_2+d_4 = \text{Even}$ implying that a divisor is even which is an obvious contradiction. Thus $d_2 = 2$. Now assume $d_3 = 4$. Note that this means $8$ must be a factor of $n$ (just check Binomial-Theorems solution to see all the possible combinations and note that we can't have a square of a prime in our prime factorization). If $d_4=8$ this implies $10d_6 = d_{d_5}$ implying $5$ is a factor, which is impossible. If $d_5=8$ then if $d_4 = 5$ this implies $5+2=7$ is a factor, if $d_4=6$, this implies $6=2*3$, or $3$ is a factor, and if $d_4=7$, this implies $7+2=9=3*3$ that 3 is a factor. If $d_6=8$ then we can see that the only possibility is $d_4=5$ and $d_5=7$ (as discussed above). A quick check shows that this does not work. And if we let $d_7=8$ this means we have to find $3$ primes between $4$ and $8$ even though there are only 2 => another contradiction. We know that $d_3$ has to be a prime number now, as if it was composite, then there exists a prime factor greater than $2$ in it which is less than it but greater than 2. Now, we simply mush through cases. If $d_3=3$, this implies either $d_4=5$ or $d_4=6$. If $d_4=6$ this assumes $6+2 =8$ is a factor, which it's not since $4$ is not a factor. Thus this means $d_4=5, _5=6$ and since $2+5 = 7$ we have that $d_6=7$. Using B-T's chart we see the only way we can have 4 prime factors and 16 factors is $2*3*5*7$. A quick check shows that this does not work. Thus $d_3 \ne 3$ If $d_3=5$ we see that $d_4$ is either $7$ or $10$. If $d_4=7$ then $d_2+d_4=9$ is a factor implying 3 is a factor, which it isn't. If $d_4=10$ this again implies $2+10 = 12$ is a factor, again implying 3 is a factor which it is not. Thus $d_3 \ne 5$. If $d_1=1, d_2 =2, d_3=7$ then $d_4=11,13,$ or $14$. If $d_4=13$ this implies $13+2=15$ is a factor, implying 3 and 5 are factors which they are not. If $d_4=14$ this implies $14+2 = 16$ is a factor, implying false factors of $4$ and $8$. Thus the only possibility is $d_4=11$ which implies $d_5= 13$. Again, the only way to have these 4 prime factors is $2*7*11*13$. A quick check shows this works. Thus one solution is $2002$. If $d_1=1, d_2=2, d_3=11$, then $d_4= 13,17,19,22$. As before, we quickly show $d_4=13,19,22$ don't work, making the only possibility $d_4=17$. But we have the restriction that $d_5 \le 16$ (or else $d_{d_5}$ isn't a factor of $n$). Since $d_5 > d_4 =17$ this is a contradiction. Finally, $d_1=1, d_2=2, d_3=13$. The next prime number of $13$ is $17 \ge 16$. Thus this means that $d_5 > 17$ which doesn't work. And subsequent primes will not work either. Thus the only solution is 2002. Sorry this was rushed I may come back and edit given that I have to go somewhere around right now

Valentin Vornicu wrote: Find all positive integers which have exactly 16 positive divisors $1 = d_1 < d_2 < \ldots < d_{16} =n$ such that the divisor $d_k$, where $k = d_5$, equals $(d_2 + d_4) d_6$. Let $n=\prod_{i=1}^{a} (p_i)^{b_i}$ then No. Of divisors of $n=\prod_{i=1}^{a}(1+b_i)=16$ So $1\leq a \leq 4$ Hence there are some cases.. Case 1-: when $a=1$ then $n=(p_1)^{15}$ then $d_5=(p_1)^4$ and as $d_{d_5}=d_6(d_4+d_2) \implies d_{(p_1)^4}=(p_1)^6(p_1^3+1)$ which is not possible. Case 2-: when $a=2$ then $n=p_1^{b_1}*p_2^{b_2}$ Then $(1+b_1)(1+b_2)=16=2*8=4*4$ 1-: let $b_1=1, b_2=7$ then $n=p_1*p_2^7$ and let $p_1>p_2$ then $d_2=p_2, d_4=p_2^2$ and as $d_2+d_4|n \implies p_2(p_2+1)|n$ which is not possible. 2-: when $b_1=b_2=3$ then $n=p_1^3*p_2^3$ and let $p_1>p_2$ then again $d_2=p_2, d_4=p_2^2$ and as $d_2+d_4|n \implies p_2(p_2+1)|n$ which is not possible. So no solution in this case. Case 3-: when $a=3$ then $n=p_1^{b_1}*p_2^{b_2}*p_3^{b_3}$ then $(1+b_1)(1+b_2)(1+b_3)=16=2*2*4$ 1-: when $b_1=b_2=1, b_3=3$ then $n=p_1*p_2*p_3^3$ and atleast 1 prime divisor should be even then let $p_1=2$ and $p_2>p_3$ Then $d_2=2, d_4=p_2, d_5=2*p_3, d_6=2*p_2$ Then $6<2*p_3\leq 16 \implies p_3=5$ or $7$ And also $2*p_2(2+p_2)|2*p_2*p_3^3\implies 2+p_2=p_3^2$ if $p_3=5$ then $p_2=23$ but Then $d_6=p_3^2=25$ hence $p_3\neq 5$ If $p_3=7$ then $p_2=47$ but then $d_5=p_2=46$ hence $p_3\neq 7$. Hence no solution in this case. Case 4-: when $a=4$ clearly then $b_1=b_2=b_3=b_4=1$ and $n=2*p_2*p_3*p_4$ WLOG $p_1>p_2>p_3$ 1-: then If $d_2=2, d_4=p_2, d_5=p_1, d_6=2*p_3$ Then $2*p_3(p_2+2)|2*p_2*p_3*p_1$ which is only possible if $p_2+2=p_1$ and also $6<p_1\leq 16$ possible values of $p_1\in \{7, 11,13\}$ If $p_1=7$ then $p_3=3, p_2=5$ but then $d_5\neq 7$ so $p_1\neq 7$ If $p_1=11$ then also $p_2=5, p_3=3$ is not possible. So checking $p_2=7, p_3=5$ gives $d_5=10$ not $11$ so $p_1\neq 11$. If $p_1=13$ then $p_2=11, p_3=7$ works as $2*p_2>p_1$ so $d_5=p_1$. Hence $\boxed{n=2002}$ is a solution. And now still some subcases of case 4 is left which is when $2*p_i<p_k$ for $i, k\leq 4$ which can easily be solved by above reasoning and gives No solution. So $\boxed{n=2002}$ is the only solution.