<s>[INST] How many N-digit numbers have the sum of odd places digits equal the sum of even places digits? We define a N-digit number as a sequence $d_{1}d_{2}d_3...d_N$, where $d_k\in\{0,1,2,3,4,5,6,7,8,9\}$. So the first digits can be equal to zero. N is a positive even integer.
The problem is to find, how many N-digit numbers are there, which have their sum of the digits on odd places equal to the sum of digits on the even places. Example number would be 2563 for 4-digit numbers as 2 + 6 = 5 + 3.
My thought was to choose the odd digits, we can do that in $10^{N/2}$ ways and then for each of that combinations make even digits a permutation of them ($\frac{N}{2}!$), but this, if correct at all, is only a small fraction of all the possibilities. Why? Because the digits do not have to be the same. Only the sums have to be equal.
Then my second though was that each of such numbers consists of digits from other (shorter) numbers that satisfy the requirement, but it is not true since only the sums have to be equal.
We can for each sum of odd/even digits calculate how many ways we can express it on N/2 digits and then sum the squares of such numbers of combinations. However that leads to very unfriendly calculations.
After you find a nice way, please provide some example calculations for N=4 and N=6.
 [/INST] This answer is intended to be simple and self-contained but please ask if it has not been expressed sufficiently clearly.
The expansion of $(1+x+x^2+...+x^9)^n$ as $\sum a_ix^i$
Consider multiplying out  $(1+x+x^2+...+x^9)(1+x+x^2+...+x^9)...(1+x+x^2+...+x^9).$
For a given integer $M$, a term $a_Mx^M$ in the product is given by the sum of terms $$x^ux^v...$$
where $x^u$ is from the first bracket, $x^v$ is from the second bracket and so on. Thus $a_M$ is precisely the number of ways of expressing the number $M$ as the sum of $n$ digits.
The sum of squares
We require $\sum a_M^2$. This is the constant coefficient in the product $$\sum a_Mx^M\sum a_Mx^{-M}$$ or, equivalently, the coefficient of $x^{9n}$ in $(1+x+x^2+...+x^9)^{2n}$
The calculation for $N=4$ i.e. n=2
We require the coefficient of $x^{18}$ in $(1+x+x^2+...+x^9)^{4}=(1-x^{10})^4(1-x)^{-4}$.
Expanding, we obtain
$$(1-4x^{10}+...)(1+4x+...+\begin{pmatrix}11\\8\\\end{pmatrix}x^{8}+...+\begin{pmatrix}21\\18\\\end{pmatrix}x^{18}+...)$$
The required number is $\begin{pmatrix}21\\18\\\end{pmatrix}-4\begin{pmatrix}11\\8\\\end{pmatrix}=670.$
The calculation for $N=6$ i.e. n=3
We now require the coefficient of $x^{27}$ in $(1-x^{10})^6(1-x)^{-6}$. This is
$$(1-6x^{10}+15x^{20}-...)(1...+\begin{pmatrix}12\\7\\\end{pmatrix}x^{7}+...+\begin{pmatrix}22\\17\\\end{pmatrix}x^{17}+...\begin{pmatrix}32\\27\\\end{pmatrix}x^{27}+...)$$
The required number is $\begin{pmatrix}32\\27\\\end{pmatrix}-6\begin{pmatrix}22\\17\\\end{pmatrix}+15\begin{pmatrix}12\\7\\\end{pmatrix}=55252.$
A general formula
Let $K=9N/2$. Then the calculation we require is just $$\begin{pmatrix}K+N-1\\K\\\end{pmatrix}-\begin{pmatrix}N\\1\\\end{pmatrix}\begin{pmatrix}K+N-11\\K-10\\\end{pmatrix}+\begin{pmatrix}N\\2\\\end{pmatrix}\begin{pmatrix}K+N-21\\K-20\\\end{pmatrix}-....$$
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