Let's determine the value of $n$, given that

$$
\binom{n}{5}=\frac{n(n-1)(n-2)(n-3)(n-4)}{2 \cdot 3 \cdot 4 \cdot 5}
$$

(which, as we know, is an integer) is of the form $\overline{a b a b a b}$ in the decimal system, where $a$ and $b$ represent digits.
In the decimal system, a number of the form ababab can be written as: $a \cdot 101010 + b \cdot 10101 = (10a + b) 10101 : \binom{n}{5}$, so it is divisible by 10101. Factoring 10101 into prime factors: $10101 = 3 \cdot 7 \cdot 13 \cdot 37$. On the other hand, according to the interpretation of $\binom{n}{5}$, $(n-1)(n-2)(n-3)(n-4) = 5! \cdot \binom{n}{5}$, so $n(n-1)(n-2)(n-3)(n-4)$ is divisible by $3 \cdot 7 \cdot 13 \cdot 37$. Therefore, one of the numbers $n, n-1, n-2, n-3, n-4$ must be divisible by 37; a multiple of 37 larger than 2 \cdot 37 cannot appear among the five numbers because then $\binom{n}{5}$ would already be a number with more than 6 digits.

If $2 \cdot 37$ appears among the factors, then $n$ is at least 70, and thus $\binom{n}{5} \geq \binom{70}{5} = \frac{70 \cdot 69 \cdot 68 \cdot 67 \cdot 66}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} > \frac{66 \cdot 66 \cdot 66 \cdot 66 \cdot 66}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} > 10436104$, but this is impossible because ababab < 1000000. Therefore, one of the numbers is 37.

Among five consecutive numbers, one must also be divisible by 13. Let's find the number closest to 37 that is divisible by 13, which is 39. Other multiples of 13 are more than 5 units away from 37, so the three numbers among the five are: 39, 38, 37. The product of these three numbers is also divisible by 3, so one of the remaining two numbers must be divisible by 7. Among the numbers 41, 40, 36, and 35, only one is divisible by 7: 35. Therefore, the five numbers are: 39, 38, 37, 36, 35.

Thus, if the problem has a solution, it can only be $n = 39$, and this is indeed a solution because $\binom{39}{5} = 575757$.