Consider the fourteen numbers, $$1^4, 2^4, \dots, 14^4$$. The smallest natural number $$n$$ such that they leave distinct remainders when divided by $$n$$ is:

We need the least natural number $$n$$ for which the fourteen fourth-powers
$$1^4,\,2^4,\,3^4,\,\dots ,\,14^4$$
give fourteen different remainders on division by $$n$$. In other words, for every pair $$i\neq j$$ in $$\{1,2,\dots ,14\}$$ we must have $$n \nmid i^4-j^4$$.

Step 1 : A trivial lower bound
There are fourteen numbers. If all remainders are to be different, we obviously need $$n\ge 14$$ (otherwise the pigeon-hole principle forces a repetition).

Step 2 : The range $$15\le n\le 28$$ always fails
For any such $$n$$ let $$a=n-14$$. Because $$15\le n\le 28$$ we get $$1\le a\le 14$$ and $$a\neq 14$$.
Now observe the identity$$(n-a)^4\equiv a^4\pmod n.\tag{-1}$$Choosing $$a=n-14$$ gives the pair $$a$$ and $$14$$ inside the set $$\{1,\dots ,14\}$$ with the same remainder by (-1).
Hence no $$n$$ in this interval can work.

Step 3 : The value $$n=14$$ also fails
A short table suffices:

$$\begin{array}{c|cccccccccccccc} k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14\\\hline k^4\bmod 14 & 1 & 2 & 11 & 4 & 9 & 8 & 7 & 8 & 9 & 4 & 11 & 2 & 1 & 0 \end{array}$$
The remainder $$8$$ appears for both $$k=6$$ and $$k=8$$, so $$n=14$$ is ruled out.

Step 4 : Eliminate $$n=29$$ and $$n=30$$

• $$n=29$$ (prime)
Compute a few fourth powers modulo 29: $$1^4\equiv1,\;2^4\equiv16,\;5^4\equiv16\pmod{29}.$$ The pair $$2,5$$ already coincid​e, so 29 is impossible.

• $$n=30$$
Because $$30=2\cdot3\cdot5,$$ equal remainders mod 30 appear as soon as they coincide modulo any prime factor. Indeed, $$1^4\equiv1,\;7^4=2401\equiv1\pmod{30}$$, giving a repetition. Hence 30 is impossible.

Step 5 : Show that $$n=31$$ works

31 is prime and $$31\equiv3\pmod4$$. In such a prime field, the equation$$x^4\equiv y^4\pmod{31}$$implies $$x\equiv\pm y\pmod{31}$$ for all non-zero residues, because the remaining factor $$x^2+y^2$$ cannot vanish (−1 is a quadratic non-residue when the prime is 3 mod 4).

Among the integers $$1,2,\dots ,14$$ no two are negatives of each other modulo 31 (their negatives lie in $$17,\dots ,30$$). Therefore their fourth powers are all distinct modulo 31.

Thus 31 is admissible, and every smaller $$n\ge14$$ has already been excluded.

Final answer: the smallest such natural number is
$$\boxed{31}$$.