The integers $a_{1}, a_{2}, \ldots, a_{10}$ are greater than 1 and their sum is 2006. What is the smallest possible value of

$$
\binom{a_{1}}{2}+\ldots+\binom{a_{10}}{2}
$$

sum?
Solution. The sum of positive integers can only be 2006 in a finite number of ways, so the desired sum has a minimum. Let's write the sum in the following form:

$$
\binom{a_{1}}{2}+\ldots+\binom{a_{10}}{2}=\frac{a_{1}^{2}-a_{1}}{2}+\ldots+\frac{a_{10}^{2}-a_{10}}{2}=\frac{a_{1}^{2}+\ldots a_{10}^{2}}{2}-1003
$$

From this, it is clear that it is sufficient to find the smallest value of the sum of squares. Let six of the $a_{i}$ numbers be 201, and the other four be 200; this is the only distribution that satisfies the conditions of the problem, where the difference between any two of the ten numbers is at most 1. We will show that in any other case, the numbers can be changed within the constraints of the conditions so that their sum of squares decreases. In such a case, there will always be two of them, $a$ and $b$, for which, for example, $a \geq b+2$; replace $a$ with $(a-1)$ and $b$ with $(b+1)$. Then

$$
\left(a^{2}+b^{2}\right)-\left((a-1)^{2}+(b+1)^{2}\right)=2(a-b-1) \geq 2
$$

Therefore, the sum of squares can only take its minimum value in the above case, which is $6 \cdot 201^{2}+4 \cdot 200^{2}=402406$, so the smallest value of the sum in the problem is

$$
\frac{402406}{2}-1003=200200
$$