Problem 11. Denote by $d(a, b)$ the number of the divisors of a positive integer $a$, which are greater than or equal to $b$. Find all positive integers $n$ such that

$$
d(3 n+1,1)+d(3 n+2,2)+\cdots+d(4 n, n)=2006
$$

Ivan Landjev
11. Denote by $D(a, b)$ the set of the divisors of $a$, which are greater than or equal to $b$. Thus, $|D(a, b)|=d(a, b)$. Every integer $k, 1 \leq k \leq 4$, belongs to at most one of the sets

$$
D(3 n+1,1), D(3 n+2,2), \ldots, D(4 n, n)
$$

Every integer $k, 1 \leq k \leq n, 3 n+1 \leq k \leq 4 n$ belongs to exactly one of the sets (1). The integers $k, 2 n+1 \leq k \leq 3 n$ do not appear in the sets (1).

Let $n+1 \leq k \leq 2 n$, i.e. $k=n+i, i=1, \ldots, n$. If $k$ belongs to one of the sets (1), then

$$
3 n+1 \leq 2(n+i) \leq 4 n
$$

or

$$
3 n+1 \leq 3(n+i) \leq 4 n
$$

We conclude that $i=\left\lceil\frac{n+1}{2}\right\rceil, \ldots, n$ or $i=1, \ldots,\left\lfloor\frac{n}{3}\right\rfloor$. The number of the integers from the interval $[n+1,2 n]$ that belong to exactly one of the sets (1) equals $\left\lceil\frac{n}{2}\right\rceil+\left\lfloor\frac{n}{3}\right\rfloor$. Thus

$$
|D(3 n+1,1)|+|D(3 n+2,2)|+\cdots+|D(4 n, n)|=2 n+\left\lceil\frac{n}{2}\right\rceil+\left\lfloor\frac{n}{3}\right\rfloor
$$

and therefore $n=708$.