4. Determine all natural numbers $n$ that have the following properties: $n$ is divisible by 2 but not by 4, the sum of the digits of $n$ is equal to 6, the number of divisors of $n$ is equal to 8, the sum of the divisors of $n$ is divisible by 10, and $n$ does not give a remainder of 12 or 14 when divided by 16.
4. It is known that $n$ is divisible by 2, and since the sum of its digits is divisible by 3, $n$ must also be divisible by 3. We also know that $n$ is not divisible by 4, and also not by 9 (since the sum of its digits is not divisible by 9), so $n$ must have another prime factor, as it has exactly 8 divisors. From this, we conclude that $n$ must be of the form $n=6p$, where $p$ is a prime number different from 2 and 3. Now we can determine the sum of the divisors of $n$: it is $1+2+3+6+p+2p+3p+6p=12(p+1)$, and since this number is divisible by 10 according to the problem, it must be that $5 \mid p+1$. From this, the prime number $p$ ends in the digit 9, and the number $n=6p$ ends in the digit 4. The sum of the digits of $n$ is 6, which leaves three possibilities for the two-digit ending of the number: 04, 14, and 24. However, if the number ends in 04 or 24, it is divisible by 4, which is not true for $n$, so the two-digit ending of the number $n$ is 14. Since the sum of the digits of $n$ is 6, it follows that $n=10^x+14$ for some natural number $x>1$. For $x \geqslant 4$, $n \equiv 14(\bmod 16)$, which is excluded by the condition in the problem. Therefore, we need to examine the possibilities $x=2$ and $x=3$. In the case $x=2$, we get the solution $114=6 \cdot 19$ (which gives a remainder of 2 when divided by 16), while in the case $x=3$, there is no solution because $1014=6 \cdot 169$ and 169 is not a prime number (which contradicts the form $n=6p$). Therefore, the only solution is $n=114$.