Let $m_1 < m_2 < \dots < m_n$ be the number of points of each colour. We call $m_1, m_2, \dots, m_n$ the colour distribution. Then $m_1 + \dots + m_n = 2012$ and the number of multi-coloured sets is $M = m_1 m_2 \dots m_n$. We have the following observations.

(i) $m_1 > 1$. For if $m_1 = 1$, then $m_1 m_2 \dots m_n < m_2 m_3 \dots m_{n-1} (1 + m_n)$. This means if we use $n-1$ colours with colour distribution $m_2, m_3, \dots, m_{n-1}, (1+m_n)$, we obtain a larger $M$.

(ii) $m_{i+1} - m_i \le 2$ for all $i$. For if there exists $k$ with $m_{k+1} - m_k \ge 3$, then the colour distribution with $m_k, m_{k+1}$ replaced by $m_k + 1, m_{k+1} - 1$ yields a larger $M$.

(iii) $m_{i+1} - m_i = 2$ for at most one $i$. For if there exist $i < j$ with $m_{i+1} - m_i = m_{j+1} - m_j = 2$, the colour distribution with $m_i, m_{j+1}$ replaced by $m_i + 1, m_{j+1} - 1$ yields a larger $M$.

(iv) $m_{i+1} - m_i = 2$ for exactly one $i$. For if $m_{i+1} - m_i = 1$ for all $i$, then $m_1 + \dots + m_n = n m_1 + \frac{n(n-1)}{2} = 2012 = 4 \cdot 503$. Thus $n(2m_1 - 1 + n) = 8 \cdot 503$. Since $503$ is prime, the parity of $n$ and $2m_1 - 1 + n$ are opposite and $2m_1 - 1 + n > n$, we have $n=8$ and $m_1 = 248$. The colour distribution with $m_1$ replaced by two numbers $2, 246$ (using $n+1$ colours) yields a larger $M$.

(v) $m_1 = 2$. If $m_n - m_{n-1} = 2$, then from (iv), we have $m_1 + \dots + m_n = n m_1 + \frac{n(n-1)}{2} + 1 = 2012$. Thus $n(2m_1 - 1 + n) = 2 \cdot 2011$. Since $2011$ is prime, we get $n=2$ and $m_1 = 1005$ which will lead to a contradiction as in (iv). Thus $m_n - m_{n-1} = 1$. $m_{i+1} - m_i = 2$ for some $1 \le i \le n-2$. Suppose $m_1 \ge 3$. Let $m' = m_{i+2} - 2$. Then $m_i < m' < m_{i+1}$ with replacing $m_{i+2}$ by $2, m'$ yields a larger $M$. Thus $m_1 = 2$.

From the above analysis, with $n$ colours, we see that the colour distribution $2, 3, \dots, i-1, i+1, i+2, \dots, n+1, n+2$, with $3 \le i \le n$, yields the maximum $M$. Now we have $\sum m_i = \frac{1}{2}(n+1)(n+4) - i = 2012$. Thus $n^2 + 5n - 4020 = 2i$, $3 \le i \le n$, i.e., $n^2 + 5n \ge 4026$ and $n^2 + 3n \le 4020$. Thus $n=61$ and $i=3$. Thus the maximum is achieved when $n=61$ with the colour distribution $2, 4, 5, 6, \dots, 63$.