It is possible if and only if 4 divides n. Douglas Zare showed that if 4 does not divide n, then it is impossible (btw, his argument might be simpler to state mod 2 - if 4 does not divide n, then it is already impossible to get all numbers divisible by 4). Now, if n=4k, do the following:

1, 12, 2, 10, 4, 8, 6, 7, 5, 9, 3, 11.

This will give 2k-1 n's, 2k-1 n+2's and two n+1's. Next, do

12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 13, 13.

This gives 4k-2 2n+2's, one 2n+3 and one 2n+1.

Now, for simplicity, subtract 2n+2 from each number. We have all zeros but one +1 and one -1. Make the permutation so that we double the number of +1's and -1's in each step, until they become >n/4, then "mix them a bit"* so that in the next step we can exactly n/2 of both. Then an alternating sequence will give all equal numbers.

$^*Edit$: As Douglas pointed out this mixing is not that clear how to do. So instead, I claim that we for any i from 1 to n/2 it is possible to get exactly i +1's and i -1's. The proof is by induction - we either double or double -1, by putting a +1 and a -1 next to each other. Eventually we can get n/2 of each and we are done.

However, this my method only works for the numbers 1, 2, .. , n. What if we start from another sequence? Mod 2 the problem is always solvable, maybe that helps, but over Z already 1, 0, .. , 0 is not clear how to solve.

It is possible if and only if 4 divides n. Douglas Zare showed that if 4 does not divide n, then it is impossible (btw, his argument might be simpler to state mod 2 - if 4 does not divide n, then it is already impossible to get all numbers divisible by 4). Now, if n=4k, do the following:

1, 12, 2, 10, 4, 8, 6, 7, 5, 9, 3, 11.

This will give 2k-1 n's, 2k-1 n+2's and two n+1's. Next, do

12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 13, 13.

This gives 4k-2 2n+2's, one 2n+3 and one 2n+1.

Now, for simplicity, subtract 2n+2 from each number. We have all zeros but one +1 and one -1. Make the permutation so that we double the number of +1's and -1's in each step, until they become >n/4, then "mix them a bit"* so that in the next step we can exactly n/2 of both. Then an alternating sequence will give all equal numbers.

$^*Edit$: As Douglas pointed out this mixing is not that clear how to do. So instead, I claim that we for any i from 1 to n/2 it is possible to get exactly i +1's and i -1's. The proof is by induction - we either double or double -1, by putting a +1 and a -1 next to each other. Eventually we can get n/2 of each and we are done.

However, this my method only works for the numbers 1, 2, .. , n. What if we start from another sequence? Mod 2 the problem is always solvable, maybe that helps, but over Z already 1, 0, .. , 0 is not clear how to solve.

Edit: As Doublas pointed out this is possible if and only if n is a power of 2. So to summarize, I think if 4 does not divide n>1, then you can not get all equal numbers starting from 1, 2, .. , n, if 4 divides n you can, while if is n>2 is a power of 2, then it seems you can start from any sequence. (Though we have not seen a full proof of this last part yet.)

It is possible if and only if 4 divides n. Douglas Zare showed that if 4 does not divide n, then it is impossible (btw, his argument might be simpler to state mod 2 - if 4 does not divide n, then it is already impossible to get all numbers divisible by 4). Now, if n=4k, do the following:

1, 12, 2, 10, 4, 8, 6, 7, 5, 9, 3, 11.

This will give 2k-1 n's, 2k-1 n+2's and two n+1's. Next, do

12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 13, 13.

This gives 4k-2 2n+2's, one 2n+3 and one 2n+1.

Now, for simplicity, subtract 2n+2 from each number. We have all zeros but one +1 and one -1. Make the permutation so that we double the number of +1's and -1's in each step, until they become >n/4, then mix them a bit so that in the next step we can exactly n/2 of both. Then an alternating sequence will give all equal numbers.

However, this my method only works for the numbers 1, 2, .. , n. What if we start from another sequence? Mod 2 the problem is always solvable, maybe that helps, but over Z already 1, 0, .. , 0 is not clear how to solve.

It is possible if and only if 4 divides n. Douglas Zare showed that if 4 does not divide n, then it is impossible (btw, his argument might be simpler to state mod 2 - if 4 does not divide n, then it is already impossible to get all numbers divisible by 4). Now, if n=4k, do the following:

1, 12, 2, 10, 4, 8, 6, 7, 5, 9, 3, 11.

This will give 2k-1 n's, 2k-1 n+2's and two n+1's. Next, do

12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 13, 13.

This gives 4k-2 2n+2's, one 2n+3 and one 2n+1.

Now, for simplicity, subtract 2n+2 from each number. We have all zeros but one +1 and one -1. Make the permutation so that we double the number of +1's and -1's in each step, until they become >n/4, then "mix them a bit"* so that in the next step we can exactly n/2 of both. Then an alternating sequence will give all equal numbers.

$^*Edit$: As Douglas pointed out this mixing is not that clear how to do. So instead, I claim that we for any i from 1 to n/2 it is possible to get exactly i +1's and i -1's. The proof is by induction - we either double or double -1, by putting a +1 and a -1 next to each other. Eventually we can get n/2 of each and we are done.

However, this my method only works for the numbers 1, 2, .. , n. What if we start from another sequence? Mod 2 the problem is always solvable, maybe that helps, but over Z already 1, 0, .. , 0 is not clear how to solve.